Python data analysis Learning note ch05

pandas入门

按照以下约定引用相关package

from pandas import Series, DataFrame
import pandas as pd

from __future__ import division
from numpy.random import randn
import numpy as np
import os
import matplotlib.pyplot as plt
np.random.seed(12345)
plt.rc('figure', figsize=(10, 6))
from pandas import Series, DataFrame
import pandas as pd
np.set_printoptions(precision=4)
from IPython.core.interactiveshell import InteractiveShell
InteractiveShell.ast_node_interactivity = "all"

pandas数据结构介绍

Series

Series是一种类似于一维数组的对象，由一组数据以及一组与之相关的数据标签（类似于字典的键）组成，所以可以看成是一个有序的字典

obj = Series([4, 7, -5, 3])
obj

0    4
1    7
2   -5
3    3
dtype: int64

obj.values
obj.index

array([ 4,  7, -5,  3], dtype=int64)






RangeIndex(start=0, stop=4, step=1)

obj2 = Series([4, 7, -5, 3], index=['d', 'b', 'a', 'c'])
obj2

d    4
b    7
a   -5
c    3
dtype: int64

obj2.index

Index(['d', 'b', 'a', 'c'], dtype='object')

obj2['a']

-5

obj2['d'] = 6
obj2[['c', 'a', 'd']]

c    3
a   -5
d    6
dtype: int64

各种Numpy运算都是作用在数据上，同时索引与数据的链接会一直保持

obj2[obj2 > 0]

d    6
b    7
c    3
dtype: int64

obj2 * 2

d    12
b    14
a   -10
c     6
dtype: int64

np.exp(obj2)

d     403.428793
b    1096.633158
a       0.006738
c      20.085537
dtype: float64

'b' in obj2

True

'e' in obj2

False

因此可以直接根据Numpy的Dict来创建Series

sdata = {'Ohio': 35000, 'Texas': 71000, 'Oregon': 16000, 'Utah': 5000}
obj3 = Series(sdata)
obj3

Ohio      35000
Oregon    16000
Texas     71000
Utah       5000
dtype: int64

如果传入了index参数的话，那么就会与传入的Dict做键匹配，没有匹配上的就设为NaN

states = ['California', 'Ohio', 'Oregon', 'Texas']
obj4 = Series(sdata, index=states)
obj4

California        NaN
Ohio          35000.0
Oregon        16000.0
Texas         71000.0
dtype: float64

pd.isnull(obj4)

California     True
Ohio          False
Oregon        False
Texas         False
dtype: bool

pd.notnull(obj4)

California    False
Ohio           True
Oregon         True
Texas          True
dtype: bool

obj4.isnull()

California     True
Ohio          False
Oregon        False
Texas         False
dtype: bool

Series有一个非常重要的数据对齐的功能

obj3

Ohio      35000
Oregon    16000
Texas     71000
Utah       5000
dtype: int64

obj4

California        NaN
Ohio          35000.0
Oregon        16000.0
Texas         71000.0
dtype: float64

obj3 + obj4

California         NaN
Ohio           70000.0
Oregon         32000.0
Texas         142000.0
Utah               NaN
dtype: float64

Series本身以及其索引都有一个叫做name的属性，这个属性十分重要，以后很多高级功能都会用到

obj4.name = 'population'
obj4.index.name = 'state'
obj4

state
California        NaN
Ohio          35000.0
Oregon        16000.0
Texas         71000.0
Name: population, dtype: float64

可以通过直接赋值的方式修改index属性

obj.index = ['Bob', 'Steve', 'Jeff', 'Ryan']
obj

Bob      4
Steve    7
Jeff    -5
Ryan     3
dtype: int64

DataFrame

DataFrame是一个表格型的数据结构，可以看成由Series组成的字典，只不过这些Series共用一套索引

data = {'state': ['Ohio', 'Ohio', 'Ohio', 'Nevada', 'Nevada'],
        'year': [2000, 2001, 2002, 2001, 2002],
        'pop': [1.5, 1.7, 3.6, 2.4, 2.9]}
frame = DataFrame(data)

数据会被排序

frame

	pop	state	year
0	1.5	Ohio	2000
1	1.7	Ohio	2001
2	3.6	Ohio	2002
3	2.4	Nevada	2001
4	2.9	Nevada	2002

DataFrame(data, columns=['year', 'state', 'pop'])

	year	state	pop
0	2000	Ohio	1.5
1	2001	Ohio	1.7
2	2002	Ohio	3.6
3	2001	Nevada	2.4
4	2002	Nevada	2.9

frame2 = DataFrame(data, columns=['year', 'state', 'pop', 'debt'],
                   index=['one', 'two', 'three', 'four', 'five'])
frame2

	year	state	pop	debt
one	2000	Ohio	1.5	NaN
two	2001	Ohio	1.7	NaN
three	2002	Ohio	3.6	NaN
four	2001	Nevada	2.4	NaN
five	2002	Nevada	2.9	NaN

frame2.columns

Index(['year', 'state', 'pop', 'debt'], dtype='object')

DataFrame每一个Key对应的Value都是一个Series

frame2['state']

one        Ohio
two        Ohio
three      Ohio
four     Nevada
five     Nevada
Name: state, dtype: object

frame2.year

one      2000
two      2001
three    2002
four     2001
five     2002
Name: year, dtype: int64

注意到name属性也已经被设置好了

ix相当于一个行索引？

frame2.ix['three']

year     2002
state    Ohio
pop       3.6
debt      NaN
Name: three, dtype: object

可以利用Numpy的广播功能

frame2['debt'] = 16.5
frame2

	year	state	pop	debt
one	2000	Ohio	1.5	16.5
two	2001	Ohio	1.7	16.5
three	2002	Ohio	3.6	16.5
four	2001	Nevada	2.4	16.5
five	2002	Nevada	2.9	16.5

也可以赋值一个列表，但是长度必须匹配

frame2['debt'] = np.arange(5.)
frame2

	year	state	pop	debt
one	2000	Ohio	1.5	0.0
two	2001	Ohio	1.7	1.0
three	2002	Ohio	3.6	2.0
four	2001	Nevada	2.4	3.0
five	2002	Nevada	2.9	4.0

如果是赋值一个Series，则会匹配上索引，没有匹配上的就是置为NaN

val = Series([-1.2, -1.5, -1.7], index=['two', 'four', 'five'])
frame2['debt'] = val
frame2

	year	state	pop	debt
one	2000	Ohio	1.5	NaN
two	2001	Ohio	1.7	-1.2
three	2002	Ohio	3.6	NaN
four	2001	Nevada	2.4	-1.5
five	2002	Nevada	2.9	-1.7

也可以进行逻辑操作

frame2['eastern'] = frame2.state == 'Ohio'
frame2

	year	state	pop	debt	eastern
one	2000	Ohio	1.5	NaN	True
two	2001	Ohio	1.7	-1.2	True
three	2002	Ohio	3.6	NaN	True
four	2001	Nevada	2.4	-1.5	False
five	2002	Nevada	2.9	-1.7	False

del用于删除一列

del frame2['eastern']
frame2.columns

Index(['year', 'state', 'pop', 'debt'], dtype='object')

只要是通过索引方式进行的操作，都是直接在原数据上进行的操作，不是一个副本

嵌套的字典也可直接生成DataFrame，只不过内层的键被当作index，外层的键被当作colums

pop = {'Nevada': {2001: 2.4, 2002: 2.9},
       'Ohio': {2000: 1.5, 2001: 1.7, 2002: 3.6}}

frame3 = DataFrame(pop)
frame3

	Nevada	Ohio
2000	NaN	1.5
2001	2.4	1.7
2002	2.9	3.6

同样可以进行转置，这样的话index和column就会互换

frame3.T

	2000	2001	2002
Nevada	NaN	2.4	2.9
Ohio	1.5	1.7	3.6

显式地指定索引，不匹配的会置为NaN

DataFrame(pop, index=[2001, 2002, 2003])

	Nevada	Ohio
2001	2.4	1.7
2002	2.9	3.6
2003	NaN	NaN

frame3

	Nevada	Ohio
2000	NaN	1.5
2001	2.4	1.7
2002	2.9	3.6

还可以这样构建

pdata = {'Ohio': frame3['Ohio'][:-1],
         'Nevada': frame3['Nevada'][:2]}
DataFrame(pdata)

	Nevada	Ohio
2000	NaN	1.5
2001	2.4	1.7

frame3

	Nevada	Ohio
2000	NaN	1.5
2001	2.4	1.7
2002	2.9	3.6

name属性也会在表格中显示出来

frame3.index.name = 'year'; frame3.columns.name = 'state'
frame3

state	Nevada	Ohio
year
2000	NaN	1.5
2001	2.4	1.7
2002	2.9	3.6

values方法只返回数据，不返回index以及key

frame3.values

array([[ nan,  1.5],
       [ 2.4,  1.7],
       [ 2.9,  3.6]])

frame2

	year	state	pop	debt
one	2000	Ohio	1.5	NaN
two	2001	Ohio	1.7	-1.2
three	2002	Ohio	3.6	NaN
four	2001	Nevada	2.4	-1.5
five	2002	Nevada	2.9	-1.7

frame2.values

array([[2000, 'Ohio', 1.5, nan],
       [2001, 'Ohio', 1.7, -1.2],
       [2002, 'Ohio', 3.6, nan],
       [2001, 'Nevada', 2.4, -1.5],
       [2002, 'Nevada', 2.9, -1.7]], dtype=object)

索引对象

Index是一个可以单独提取出来的对象

obj = Series(range(3), index=['a', 'b', 'c'])
index = obj.index
index

Index(['a', 'b', 'c'], dtype='object')

index[1:]

Index(['b', 'c'], dtype='object')

不可修改~！

index[1] = 'd'

---------------------------------------------------------------------------

TypeError                                 Traceback (most recent call last)

<ipython-input-52-676fdeb26a68> in <module>()
----> 1 index[1] = 'd'


C:\Users\Ewan\Anaconda3\lib\site-packages\pandas\indexes\base.py in __setitem__(self, key, value)
   1243 
   1244     def __setitem__(self, key, value):
-> 1245         raise TypeError("Index does not support mutable operations")
   1246 
   1247     def __getitem__(self, key):


TypeError: Index does not support mutable operations

直接创建Index对象

index = pd.Index(np.arange(3))
obj2 = Series([1.5, -2.5, 0], index=index)
obj2.index is index

True

frame3

state	Nevada	Ohio
year
2000	NaN	1.5
2001	2.4	1.7
2002	2.9	3.6

'Ohio' in frame3.columns

True

2003 in frame3.index

False

基本功能

重新索引

obj = Series([4.5, 7.2, -5.3, 3.6], index=['d', 'b', 'a', 'c'])
obj

d    4.5
b    7.2
a   -5.3
c    3.6
dtype: float64

重排索引形成新对象

obj2 = obj.reindex(['a', 'b', 'c', 'd', 'e'])
obj2

a   -5.3
b    7.2
c    3.6
d    4.5
e    NaN
dtype: float64

obj.reindex(['a', 'b', 'c', 'd', 'e'], fill_value=0)

a   -5.3
b    7.2
c    3.6
d    4.5
e    0.0
dtype: float64

obj3 = Series(['blue', 'purple', 'yellow'], index=[0, 2, 4])
obj3.reindex(range(6), method='ffill') # 前向填充

0      blue
1      blue
2    purple
3    purple
4    yellow
5    yellow
dtype: object

frame = DataFrame(np.arange(9).reshape((3, 3)), index=['a', 'c', 'd'],
                  columns=['Ohio', 'Texas', 'California'])
frame

	Ohio	Texas	California
a	0	1	2
c	3	4	5
d	6	7	8

frame2 = frame.reindex(['a', 'b', 'c', 'd'])
frame2

	Ohio	Texas	California
a	0.0	1.0	2.0
b	NaN	NaN	NaN
c	3.0	4.0	5.0
d	6.0	7.0	8.0

states = ['Texas', 'Utah', 'California']
frame.reindex(columns=states)

	Texas	Utah	California
a	1	NaN	2
c	4	NaN	5
d	7	NaN	8

插值只能按行

frame.reindex(index=['a', 'b', 'c', 'd'], method='ffill',
              columns=states)

	Texas	Utah	California
a	1	NaN	2
b	1	NaN	2
c	4	NaN	5
d	7	NaN	8

用ix方法进行重新索引操作会使得代码很简洁

frame.ix[['a', 'b', 'c', 'd'], states]

	Texas	Utah	California
a	1.0	NaN	2.0
b	NaN	NaN	NaN
c	4.0	NaN	5.0
d	7.0	NaN	8.0

丢弃指定轴上的项

frame.ix[['a', 'b', 'c', 'd'], states]

	Texas	Utah	California
a	1.0	NaN	2.0
b	NaN	NaN	NaN
c	4.0	NaN	5.0
d	7.0	NaN	8.0

obj = Series(np.arange(5.), index=['a', 'b', 'c', 'd', 'e'])
new_obj = obj.drop('c')
new_obj

a    0.0
b    1.0
d    3.0
e    4.0
dtype: float64

obj.drop(['d', 'c'])

a    0.0
b    1.0
e    4.0
dtype: float64

data = DataFrame(np.arange(16).reshape((4, 4)),
                 index=['Ohio', 'Colorado', 'Utah', 'New York'],
                 columns=['one', 'two', 'three', 'four'])
data

	one	two	three	four
Ohio	0	1	2	3
Colorado	4	5	6	7
Utah	8	9	10	11
New York	12	13	14	15

data.drop(['Colorado', 'Ohio'])

	one	two	three	four
Utah	8	9	10	11
New York	12	13	14	15

data.drop('two', axis=1)

	one	three	four
Ohio	0	2	3
Colorado	4	6	7
Utah	8	10	11
New York	12	14	15

data.drop(['two', 'four'], axis=1)

	one	three
Ohio	0	2
Colorado	4	6
Utah	8	10
New York	12	14

索引，选取和过滤

多种索引方式

obj = Series(np.arange(4.), index=['a', 'b', 'c', 'd'])
obj
obj['b']

a    0.0
b    1.0
c    2.0
d    3.0
dtype: float64






1.0

obj[1]

1.0

obj[2:4]

c    2.0
d    3.0
dtype: float64

obj[['b', 'a', 'd']]

b    1.0
a    0.0
d    3.0
dtype: float64

obj[[1, 3]]

b    1.0
d    3.0
dtype: float64

obj[obj < 2] # 直接对data进行操作

a    0.0
b    1.0
dtype: float64

这种切片方式…
末端包含

obj['b':'c']

b    1.0
c    2.0
dtype: float64

obj['b':'c'] = 5
obj

a    0.0
b    5.0
c    5.0
d    3.0
dtype: float64

data = DataFrame(np.arange(16).reshape((4, 4)),
                 index=['Ohio', 'Colorado', 'Utah', 'New York'],
                 columns=['one', 'two', 'three', 'four'])
data

	one	two	three	four
Ohio	0	1	2	3
Colorado	4	5	6	7
Utah	8	9	10	11
New York	12	13	14	15

data['two']

Ohio         1
Colorado     5
Utah         9
New York    13
Name: two, dtype: int32

data[['three', 'one']]

	three	one
Ohio	2	0
Colorado	6	4
Utah	10	8
New York	14	12

data[:2] # axis=0

	one	two	three	four
Ohio	0	1	2	3
Colorado	4	5	6	7

data[data['three'] > 5]

	one	two	three	four
Colorado	4	5	6	7
Utah	8	9	10	11
New York	12	13	14	15

data < 5

	one	two	three	four
Ohio	True	True	True	True
Colorado	True	False	False	False
Utah	False	False	False	False
New York	False	False	False	False

data[data < 5] = 0

data

	one	two	three	four
Ohio	0	0	0	0
Colorado	0	5	6	7
Utah	8	9	10	11
New York	12	13	14	15

索引的另外一种选择

data.ix['Colorado', ['two', 'three']]

two      5
three    6
Name: Colorado, dtype: int32

data.ix[['Colorado', 'Utah'], [3, 0, 1]]

	four	one	two
Colorado	7	0	5
Utah	11	8	9

data.ix[2]

one       8
two       9
three    10
four     11
Name: Utah, dtype: int32

data.ix[:'Utah', 'two']

Ohio        0
Colorado    5
Utah        9
Name: two, dtype: int32

data.ix[data.three > 5, :3]

	one	two	three
Colorado	0	5	6
Utah	8	9	10
New York	12	13	14

总结一下就是说，DataFrame是一个二维的数组，只不过每一维的索引方式除了序号之外，还可以用name属性来进行索引，且一切行为与序号无异

算术运算和数据对齐

s1 = Series([7.3, -2.5, 3.4, 1.5], index=['a', 'c', 'd', 'e'])
s2 = Series([-2.1, 3.6, -1.5, 4, 3.1], index=['a', 'c', 'e', 'f', 'g'])

s1

a    7.3
c   -2.5
d    3.4
e    1.5
dtype: float64

s2

a   -2.1
c    3.6
e   -1.5
f    4.0
g    3.1
dtype: float64

数据对齐操作就是一种特殊的并集操作

s1 + s2

a    5.2
c    1.1
d    NaN
e    0.0
f    NaN
g    NaN
dtype: float64

df1 = DataFrame(np.arange(9.).reshape((3, 3)), columns=list('bcd'),
                index=['Ohio', 'Texas', 'Colorado'])
df2 = DataFrame(np.arange(12.).reshape((4, 3)), columns=list('bde'),
                index=['Utah', 'Ohio', 'Texas', 'Oregon'])
df1

	b	c	d
Ohio	0.0	1.0	2.0
Texas	3.0	4.0	5.0
Colorado	6.0	7.0	8.0

df2

	b	d	e
Utah	0.0	1.0	2.0
Ohio	3.0	4.0	5.0
Texas	6.0	7.0	8.0
Oregon	9.0	10.0	11.0

并且数据对齐操作是在所有维度上同时进行的

df1 + df2

	b	c	d	e
Colorado	NaN	NaN	NaN	NaN
Ohio	3.0	NaN	6.0	NaN
Oregon	NaN	NaN	NaN	NaN
Texas	9.0	NaN	12.0	NaN
Utah	NaN	NaN	NaN	NaN

在算术方法中填充词

下面这种定义column的方式值得注意

df1 = DataFrame(np.arange(12.).reshape((3, 4)), columns=list('abcd'))
df2 = DataFrame(np.arange(20.).reshape((4, 5)), columns=list('abcde'))
df1

	a	b	c	d
0	0.0	1.0	2.0	3.0
1	4.0	5.0	6.0	7.0
2	8.0	9.0	10.0	11.0

df2

	a	b	c	d	e
0	0.0	1.0	2.0	3.0	4.0
1	5.0	6.0	7.0	8.0	9.0
2	10.0	11.0	12.0	13.0	14.0
3	15.0	16.0	17.0	18.0	19.0

df1 + df2

	a	b	c	d	e
0	0.0	2.0	4.0	6.0	NaN
1	9.0	11.0	13.0	15.0	NaN
2	18.0	20.0	22.0	24.0	NaN
3	NaN	NaN	NaN	NaN	NaN

要想填充值必须使用add方法

df1.add(df2, fill_value=0)

	a	b	c	d	e
0	0.0	2.0	4.0	6.0	4.0
1	9.0	11.0	13.0	15.0	9.0
2	18.0	20.0	22.0	24.0	14.0
3	15.0	16.0	17.0	18.0	19.0

reindex方法与add方法还是存在差异的

df1.reindex(columns=df2.columns, fill_value=0)

	a	b	c	d	e
0	0.0	1.0	2.0	3.0	0
1	4.0	5.0	6.0	7.0	0
2	8.0	9.0	10.0	11.0	0

DataFrame和Series之间的运算

arr = np.arange(12.).reshape((3, 4))
arr

array([[  0.,   1.,   2.,   3.],
       [  4.,   5.,   6.,   7.],
       [  8.,   9.,  10.,  11.]])

arr[0]

array([ 0.,  1.,  2.,  3.])

广播操作

arr - arr[0]

array([[ 0.,  0.,  0.,  0.],
       [ 4.,  4.,  4.,  4.],
       [ 8.,  8.,  8.,  8.]])

frame = DataFrame(np.arange(12.).reshape((4, 3)), columns=list('bde'),
                  index=['Utah', 'Ohio', 'Texas', 'Oregon'])
series = frame.ix[0]
frame

	b	d	e
Utah	0.0	1.0	2.0
Ohio	3.0	4.0	5.0
Texas	6.0	7.0	8.0
Oregon	9.0	10.0	11.0

Series的name等于DataFrame的切片属性

series

b    0.0
d    1.0
e    2.0
Name: Utah, dtype: float64

默认情况下，DataFrame和Series之间的算术运算会将Series的index匹配到DataFrame的column, 然后沿着行向下广播

frame - series

	b	d	e
Utah	0.0	0.0	0.0
Ohio	3.0	3.0	3.0
Texas	6.0	6.0	6.0
Oregon	9.0	9.0	9.0

如果Series的index与DataFrame的column不匹配，则进行数据对齐

series2 = Series(range(3), index=['b', 'e', 'f'])
frame + series2

	b	d	e	f
Utah	0.0	NaN	3.0	NaN
Ohio	3.0	NaN	6.0	NaN
Texas	6.0	NaN	9.0	NaN
Oregon	9.0	NaN	12.0	NaN

series3 = frame['d']
frame

	b	d	e
Utah	0.0	1.0	2.0
Ohio	3.0	4.0	5.0
Texas	6.0	7.0	8.0
Oregon	9.0	10.0	11.0

series3

Utah       1.0
Ohio       4.0
Texas      7.0
Oregon    10.0
Name: d, dtype: float64

匹配行并且在列上进行广播， 就必须要指定axis

frame.sub(series3, axis=0)
frame.sub(series3, axis=1)

	b	d	e
Utah	-1.0	0.0	1.0
Ohio	-1.0	0.0	1.0
Texas	-1.0	0.0	1.0
Oregon	-1.0	0.0	1.0

	Ohio	Oregon	Texas	Utah	b	d	e
Utah	NaN	NaN	NaN	NaN	NaN	NaN	NaN
Ohio	NaN	NaN	NaN	NaN	NaN	NaN	NaN
Texas	NaN	NaN	NaN	NaN	NaN	NaN	NaN
Oregon	NaN	NaN	NaN	NaN	NaN	NaN	NaN

函数应用和映射

frame = DataFrame(np.random.randn(4, 3), columns=list('bde'),
                  index=['Utah', 'Ohio', 'Texas', 'Oregon'])

frame

	b	d	e
Utah	-0.204708	0.478943	-0.519439
Ohio	-0.555730	1.965781	1.393406
Texas	0.092908	0.281746	0.769023
Oregon	1.246435	1.007189	-1.296221

Numpy的元素级方法也可以应用到DataFrame上，直接把DataFrame当作二维的Numpy.array即可

np.abs(frame)

	b	d	e
Utah	0.204708	0.478943	0.519439
Ohio	0.555730	1.965781	1.393406
Texas	0.092908	0.281746	0.769023
Oregon	1.246435	1.007189	1.296221

f = lambda x: x.max() - x.min()

apply方法将函数映射到由各行或者各列形成的一维数组上

frame.apply(f) # axis=0

b    1.802165
d    1.684034
e    2.689627
dtype: float64

frame.apply(f, axis=1)

Utah      0.998382
Ohio      2.521511
Texas     0.676115
Oregon    2.542656
dtype: float64

def f(x):
    return Series([x.min(), x.max()], index=['min', 'max'])
frame.apply(f)
frame.apply(f, axis=1)

	b	d	e
min	-0.555730	0.281746	-1.296221
max	1.246435	1.965781	1.393406

	min	max
Utah	-0.519439	0.478943
Ohio	-0.555730	1.965781
Texas	0.092908	0.769023
Oregon	-1.296221	1.246435

元素级的函数映射applymap， 之所以叫这个名字是因为Series有一个元素级的映射函数map

format = lambda x: '%.2f' % x
frame.applymap(format)

	b	d	e
Utah	-0.20	0.48	-0.52
Ohio	-0.56	1.97	1.39
Texas	0.09	0.28	0.77
Oregon	1.25	1.01	-1.30

frame['e'].map(format)

Utah      -0.52
Ohio       1.39
Texas      0.77
Oregon    -1.30
Name: e, dtype: object

排序和排名

对索引或者column进行（字典）排序

obj = Series(range(4), index=['d', 'a', 'b', 'c'])
obj.sort_index()

a    1
b    2
c    3
d    0
dtype: int32

frame = DataFrame(np.arange(8).reshape((2, 4)), index=['three', 'one'],
                  columns=['d', 'a', 'b', 'c'])
frame.sort_index()

	d	a	b	c
one	4	5	6	7
three	0	1	2	3

frame.sort_index(axis=1)

	a	b	c	d
three	1	2	3	0
one	5	6	7	4

降序

frame.sort_index(axis=1, ascending=False)

	d	c	b	a
three	0	3	2	1
one	4	7	6	5

按照data进行排序

obj = Series([4, 7, -3, 2])
obj.sort_values()

2   -3
3    2
0    4
1    7
dtype: int64

在排序时，任何缺失值默认都会被放到Series的末尾

obj = Series([4, np.nan, 7, np.nan, -3, 2])
obj.sort_values()

4   -3.0
5    2.0
0    4.0
2    7.0
1    NaN
3    NaN
dtype: float64

frame = DataFrame({'b': [4, 7, -3, 2], 'a': [0, 1, 0, 1]})
frame

	a	b
0	0	4
1	1	7
2	0	-3
3	1	2

对指定index或者column进行排序

frame.sort_values(by='b')

	a	b
2	0	-3
3	1	2
0	0	4
1	1	7

或者根据multi-index亦或multi-column进行排序

frame.sort_values(by=['a', 'b'])

	a	b
2	0	-3
0	0	4
3	1	2
1	1	7

默认情况下，rank方法通过“为各组分配一个平均排名”的方式破坏平级关系。也就是说，如果有多个相同的值，则这些值的rank就是这些相同值rand的算术平均。

obj = Series([7, -5, 7, 4, 2, 0, 4, 4])
obj.rank()

0    7.5
1    1.0
2    7.5
3    5.0
4    3.0
5    2.0
6    5.0
7    5.0
dtype: float64

如果想按照一般方式排名

obj.rank(method='first')

0    7.0
1    1.0
2    8.0
3    4.0
4    3.0
5    2.0
6    5.0
7    6.0
dtype: float64

使用每个分组的最大排名

obj.rank(ascending=False, method='max')

0    2.0
1    8.0
2    2.0
3    5.0
4    6.0
5    7.0
6    5.0
7    5.0
dtype: float64

frame = DataFrame({'b': [4.3, 7, -3, 2], 'a': [0, 1, 0, 1],
                   'c': [-2, 5, 8, -2.5]})
frame

	a	b	c
0	0	4.3	-2.0
1	1	7.0	5.0
2	0	-3.0	8.0
3	1	2.0	-2.5

指定维度

frame.rank(axis=1)

	a	b	c
0	2.0	3.0	1.0
1	1.0	3.0	2.0
2	2.0	1.0	3.0
3	2.0	3.0	1.0

带有重复值的轴索引

obj = Series(range(5), index=['a', 'a', 'b', 'b', 'c'])
obj

a    0
a    1
b    2
b    3
c    4
dtype: int32

obj.index.is_unique

False

返回一个Series

obj['a']

a    0
a    1
dtype: int32

obj['c']

4

df = DataFrame(np.random.randn(4, 3), index=['a', 'a', 'b', 'b'])
df

	0	1	2
a	0.274992	0.228913	1.352917
a	0.886429	-2.001637	-0.371843
b	1.669025	-0.438570	-0.539741
b	0.476985	3.248944	-1.021228

df.ix['b']

	0	1	2
b	1.669025	-0.438570	-0.539741
b	0.476985	3.248944	-1.021228

汇总和计算描述统计

df = DataFrame([[1.4, np.nan], [7.1, -4.5],
                [np.nan, np.nan], [0.75, -1.3]],
               index=['a', 'b', 'c', 'd'],
               columns=['one', 'two'])
df

	one	two
a	1.40	NaN
b	7.10	-4.5
c	NaN	NaN
d	0.75	-1.3

df.sum() # axis=0 skipna=True

one    9.25
two   -5.80
dtype: float64

df.sum(axis=1) # skipna=True

a    1.40
b    2.60
c    0.00
d   -0.55
dtype: float64

df.mean(axis=1, skipna=False)

a      NaN
b    1.300
c      NaN
d   -0.275
dtype: float64

返回的是索引

df.idxmax()

one    b
two    d
dtype: object

df.cumsum()

	one	two
a	1.40	NaN
b	8.50	-4.5
c	NaN	NaN
d	9.25	-5.8

describe对于数值型和非数值型数据的行为不一样

df.describe()

C:\Users\Ewan\Anaconda3\lib\site-packages\numpy\lib\function_base.py:3834: RuntimeWarning: Invalid value encountered in percentile
  RuntimeWarning)

	one	two
count	3.000000	2.000000
mean	3.083333	-2.900000
std	3.493685	2.262742
min	0.750000	-4.500000
25%	NaN	NaN
50%	NaN	NaN
75%	NaN	NaN
max	7.100000	-1.300000

obj = Series(['a', 'a', 'b', 'c'] * 4)
obj
obj.describe()

0     a
1     a
2     b
3     c
4     a
5     a
6     b
7     c
8     a
9     a
10    b
11    c
12    a
13    a
14    b
15    c
dtype: object






count     16
unique     3
top        a
freq       8
dtype: object

相关系数和xi

import pandas_datareader.data as web

all_data = {}
for ticker in ['AAPL', 'IBM', 'MSFT', 'GOOG']:
    all_data[ticker] = web.get_data_yahoo(ticker)

price = DataFrame({tic: data['Adj Close']
                   for tic, data in all_data.items()})
volume = DataFrame({tic: data['Volume']
                    for tic, data in all_data.items()})

price[:10]

	AAPL	GOOG	IBM	MSFT
Date
2010-01-04	27.727039	313.062468	111.405000	25.555485
2010-01-05	27.774976	311.683844	110.059232	25.563741
2010-01-06	27.333178	303.826685	109.344283	25.406859
2010-01-07	27.282650	296.753749	108.965786	25.142634
2010-01-08	27.464034	300.709808	110.059232	25.316031
2010-01-11	27.221758	300.255255	108.906903	24.994007
2010-01-12	26.912110	294.945572	109.773245	24.828866
2010-01-13	27.291720	293.252243	109.537735	25.060064
2010-01-14	27.133657	294.630868	111.287245	25.563741
2010-01-15	26.680198	289.710772	110.841458	25.481172

volume[:10]

	AAPL	GOOG	IBM	MSFT
Date
2010-01-04	123432400	3927000	6155300	38409100
2010-01-05	150476200	6031900	6841400	49749600
2010-01-06	138040000	7987100	5605300	58182400
2010-01-07	119282800	12876600	5840600	50559700
2010-01-08	111902700	9483900	4197200	51197400
2010-01-11	115557400	14479800	5730400	68754700
2010-01-12	148614900	9742900	8081500	65912100
2010-01-13	151473000	13041800	6455400	51863500
2010-01-14	108223500	8511900	7111800	63228100
2010-01-15	148516900	10909600	8494400	79913200

price.pct_change

Signature: price.pct_change(periods=1, fill_method=’pad’, limit=None, freq=None, **kwargs)
Docstring:
Percent change over given number of periods.

Parameters

periods : int, default 1
Periods to shift for forming percent change

fill_method : str, default ‘pad’
How to handle NAs before computing percent changes

limit : int, default None
The number of consecutive NAs to fill before stopping

freq : DateOffset, timedelta, or offset alias string, optional
Increment to use from time series API (e.g. ‘M’ or BDay())

Returns

chg : NDFrame

Notes

By default, the percentage change is calculated along the stat
axis: 0, or Index, for DataFrame and 1, or minor for
Panel. You can change this with the axis keyword argument.

returns = price.pct_change()
returns.tail()

	AAPL	GOOG	IBM	MSFT
Date
2017-02-21	0.007221	0.004335	-0.002269	-0.002012
2017-02-22	0.002999	-0.001082	0.004937	-0.002016
2017-02-23	-0.004230	0.000686	0.002760	0.004040
2017-02-24	0.000952	-0.003236	-0.001651	0.000000
2017-02-27	0.001976	0.000772	-0.010753	-0.006035

returns.MSFT.corr(returns.IBM)

0.49525655865062668

returns.MSFT.cov(returns.IBM)

8.5880535146740545e-05

returns.corr()

	AAPL	GOOG	IBM	MSFT
AAPL	1.000000	0.409523	0.381374	0.388875
GOOG	0.409523	1.000000	0.402781	0.470781
IBM	0.381374	0.402781	1.000000	0.495257
MSFT	0.388875	0.470781	0.495257	1.000000

returns.cov()

	AAPL	GOOG	IBM	MSFT
AAPL	0.000269	0.000105	0.000075	0.000092
GOOG	0.000105	0.000244	0.000075	0.000106
IBM	0.000075	0.000075	0.000144	0.000086
MSFT	0.000092	0.000106	0.000086	0.000209

returns.corrwith(returns.IBM)

AAPL    0.381374
GOOG    0.402781
IBM     1.000000
MSFT    0.495257
dtype: float64

returns.corrwith(volume)

AAPL   -0.074055
GOOG   -0.009543
IBM    -0.194107
MSFT   -0.090724
dtype: float64

唯一值， 值计数以及成员资格

obj = Series(['c', 'a', 'd', 'a', 'a', 'b', 'b', 'c', 'c'])

uniques = obj.unique()
uniques

array(['c', 'a', 'd', 'b'], dtype=object)

obj.value_counts()

c    3
a    3
b    2
d    1
dtype: int64

pd.value_counts(obj.values, sort=False)

a    3
d    1
b    2
c    3
dtype: int64

mask = obj.isin(['b', 'c'])
mask

0     True
1    False
2    False
3    False
4    False
5     True
6     True
7     True
8     True
dtype: bool

obj[mask]

0    c
5    b
6    b
7    c
8    c
dtype: object

data = DataFrame({'Qu1': [1, 3, 4, 3, 4],
                  'Qu2': [2, 3, 1, 2, 3],
                  'Qu3': [1, 5, 2, 4, 4]})
data

	Qu1	Qu2	Qu3
0	1	2	1
1	3	3	5
2	4	1	2
3	3	2	4
4	4	3	4

result = data.apply(pd.value_counts).fillna(0)
result

	Qu1	Qu2	Qu3
1	1.0	1.0	1.0
2	0.0	2.0	1.0
3	2.0	2.0	0.0
4	2.0	0.0	2.0
5	0.0	0.0	1.0

处理缺失数据

string_data = Series(['aardvark', 'artichoke', np.nan, 'avocado'])
string_data

0     aardvark
1    artichoke
2          NaN
3      avocado
dtype: object

string_data.isnull()

0    False
1    False
2     True
3    False
dtype: bool

string_data[0] = None
string_data.isnull()

0     True
1    False
2     True
3    False
dtype: bool

滤除缺失数据

from numpy import nan as NA
data = Series([1, NA, 3.5, NA, 7])
data.dropna()

0    1.0
2    3.5
4    7.0
dtype: float64

data[data.notnull()]

0    1.0
2    3.5
4    7.0
dtype: float64

data = DataFrame([[1., 6.5, 3.], [1., NA, NA],
                  [NA, NA, NA], [NA, 6.5, 3.]])
cleaned = data.dropna()
data

	0	1	2
0	1.0	6.5	3.0
1	1.0	NaN	NaN
2	NaN	NaN	NaN
3	NaN	6.5	3.0

cleaned

	0	1	2
0	1.0	6.5	3.0

data.dropna(how='all')

	0	1	2
0	1.0	6.5	3.0
1	1.0	NaN	NaN
3	NaN	6.5	3.0

data[4] = NA
data

	0	1	2	4
0	1.0	6.5	3.0	NaN
1	1.0	NaN	NaN	NaN
2	NaN	NaN	NaN	NaN
3	NaN	6.5	3.0	NaN

data.dropna(axis=1, how='all')

	0	1	2
0	1.0	6.5	3.0
1	1.0	NaN	NaN
2	NaN	NaN	NaN
3	NaN	6.5	3.0

df = DataFrame(np.random.randn(7, 3))
df.ix[:4, 1] = NA; df.ix[:2, 2] = NA
df

	0	1	2
0	-0.204708	NaN	NaN
1	-0.555730	NaN	NaN
2	0.092908	NaN	NaN
3	1.246435	NaN	-1.296221
4	0.274992	NaN	1.352917
5	0.886429	-2.001637	-0.371843
6	1.669025	-0.438570	-0.539741

df.dropna(thresh=2)

	0	1	2
3	1.246435	NaN	-1.296221
4	0.274992	NaN	1.352917
5	0.886429	-2.001637	-0.371843
6	1.669025	-0.438570	-0.539741

填充缺失数据

df.fillna(0)

	0	1	2
0	-0.204708	0.000000	0.000000
1	-0.555730	0.000000	0.000000
2	0.092908	0.000000	0.000000
3	1.246435	0.000000	-1.296221
4	0.274992	0.000000	1.352917
5	0.886429	-2.001637	-0.371843
6	1.669025	-0.438570	-0.539741

df.fillna({1: 0.5, 3: -1})

	0	1	2
0	-0.204708	0.500000	NaN
1	-0.555730	0.500000	NaN
2	0.092908	0.500000	NaN
3	1.246435	0.500000	-1.296221
4	0.274992	0.500000	1.352917
5	0.886429	-2.001637	-0.371843
6	1.669025	-0.438570	-0.539741

fillna默认返回新对象，但也可以对现有对象就地修改

# always returns a reference to the filled object
_ = df.fillna(0, inplace=True)
df

	0	1	2
0	-0.204708	0.000000	0.000000
1	-0.555730	0.000000	0.000000
2	0.092908	0.000000	0.000000
3	1.246435	0.000000	-1.296221
4	0.274992	0.000000	1.352917
5	0.886429	-2.001637	-0.371843
6	1.669025	-0.438570	-0.539741

df = DataFrame(np.random.randn(6, 3))
df.ix[2:, 1] = NA; df.ix[4:, 2] = NA
df

	0	1	2
0	0.476985	3.248944	-1.021228
1	-0.577087	0.124121	0.302614
2	0.523772	NaN	1.343810
3	-0.713544	NaN	-2.370232
4	-1.860761	NaN	NaN
5	-1.265934	NaN	NaN

df.fillna(method='ffill')

	0	1	2
0	0.476985	3.248944	-1.021228
1	-0.577087	0.124121	0.302614
2	0.523772	0.124121	1.343810
3	-0.713544	0.124121	-2.370232
4	-1.860761	0.124121	-2.370232
5	-1.265934	0.124121	-2.370232

df.fillna(method='ffill', limit=2)

	0	1	2
0	0.476985	3.248944	-1.021228
1	-0.577087	0.124121	0.302614
2	0.523772	0.124121	1.343810
3	-0.713544	0.124121	-2.370232
4	-1.860761	NaN	-2.370232
5	-1.265934	NaN	-2.370232

data = Series([1., NA, 3.5, NA, 7])
data.fillna(data.mean())

0    1.000000
1    3.833333
2    3.500000
3    3.833333
4    7.000000
dtype: float64

层次索引

data = Series(np.random.randn(10),
              index=[['a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'd', 'd'],
                     [1, 2, 3, 1, 2, 3, 1, 2, 2, 3]])
data

a  1    0.332883
   2   -2.359419
   3   -0.199543
b  1   -1.541996
   2   -0.970736
   3   -1.307030
c  1    0.286350
   2    0.377984
d  2   -0.753887
   3    0.331286
dtype: float64

data.index

MultiIndex(levels=[['a', 'b', 'c', 'd'], [1, 2, 3]],
           labels=[[0, 0, 0, 1, 1, 1, 2, 2, 3, 3], [0, 1, 2, 0, 1, 2, 0, 1, 1, 2]])

data['b']

1   -1.541996
2   -0.970736
3   -1.307030
dtype: float64

data['b':'c']

b  1   -1.541996
   2   -0.970736
   3   -1.307030
c  1    0.286350
   2    0.377984
dtype: float64

data.ix[['b', 'd']]

b  1   -1.541996
   2   -0.970736
   3   -1.307030
d  2   -0.753887
   3    0.331286
dtype: float64

data[:, 2]

a   -2.359419
b   -0.970736
c    0.377984
d   -0.753887
dtype: float64

data.unstack()

	1	2	3
a	0.332883	-2.359419	-0.199543
b	-1.541996	-0.970736	-1.307030
c	0.286350	0.377984	NaN
d	NaN	-0.753887	0.331286

data.unstack().stack()

a  1    0.332883
   2   -2.359419
   3   -0.199543
b  1   -1.541996
   2   -0.970736
   3   -1.307030
c  1    0.286350
   2    0.377984
d  2   -0.753887
   3    0.331286
dtype: float64

frame = DataFrame(np.arange(12).reshape((4, 3)),
                  index=[['a', 'a', 'b', 'b'], [1, 2, 1, 2]],
                  columns=[['Ohio', 'Ohio', 'Colorado'],
                           ['Green', 'Red', 'Green']])
frame

		Ohio		Colorado
		Green	Red	Green
a	1	0	1	2
	2	3	4	5
b	1	6	7	8
	2	9	10	11

frame.index.names = ['key1', 'key2']
frame.columns.names = ['state', 'color']
frame

	state	Ohio		Colorado
	color	Green	Red	Green
key1	key2
a	1	0	1	2
	2	3	4	5
b	1	6	7	8
	2	9	10	11

frame['Ohio']

	color	Green	Red
key1	key2
a	1	0	1
	2	3	4
b	1	6	7
	2	9	10

创建MultiIndex对象复用

MultiIndex.from_arrays([['Ohio', 'Ohio', 'Colorado'], ['Green', 'Red', 'Green']],
                       names=['state', 'color'])

重排分级顺序

frame.swaplevel('key1', 'key2')

	state	Ohio		Colorado
	color	Green	Red	Green
key2	key1
1	a	0	1	2
2	a	3	4	5
1	b	6	7	8
2	b	9	10	11

frame.sortlevel(1)

	state	Ohio		Colorado
	color	Green	Red	Green
key1	key2
a	1	0	1	2
b	1	6	7	8
a	2	3	4	5
b	2	9	10	11

frame.swaplevel(0, 1).sortlevel(0)

	state	Ohio		Colorado
	color	Green	Red	Green
key2	key1
1	a	0	1	2
	b	6	7	8
2	a	3	4	5
	b	9	10	11

根据级别汇总统计

frame.sum(level='key2')

state	Ohio		Colorado
color	Green	Red	Green
key2
1	6	8	10
2	12	14	16

frame.sum(level='color', axis=1)

	color	Green	Red
key1	key2
a	1	2	1
	2	8	4
b	1	14	7
	2	20	10

使用DataFrame的列

frame = DataFrame({'a': range(7), 'b': range(7, 0, -1),
                   'c': ['one', 'one', 'one', 'two', 'two', 'two', 'two'],
                   'd': [0, 1, 2, 0, 1, 2, 3]})
frame

	a	b	c	d
0	0	7	one	0
1	1	6	one	1
2	2	5	one	2
3	3	4	two	0
4	4	3	two	1
5	5	2	two	2
6	6	1	two	3

frame2 = frame.set_index(['c', 'd'])
frame2

		a	b
c	d
one	0	0	7
	1	1	6
	2	2	5
two	0	3	4
	1	4	3
	2	5	2
	3	6	1

frame.set_index(['c', 'd'], drop=False)

		a	b	c	d
c	d
one	0	0	7	one	0
	1	1	6	one	1
	2	2	5	one	2
two	0	3	4	two	0
	1	4	3	two	1
	2	5	2	two	2
	3	6	1	two	3

frame2.reset_index()

	c	d	a	b
0	one	0	0	7
1	one	1	1	6
2	one	2	2	5
3	two	0	3	4
4	two	1	4	3
5	two	2	5	2
6	two	3	6	1

拓展话题

整数索引

ser = Series(np.arange(3.))
ser.iloc[-1]

2.0

ser

0    0.0
1    1.0
2    2.0
dtype: float64

ser2 = Series(np.arange(3.), index=['a', 'b', 'c'])
ser2[-1]

2.0

ser.ix[:1]

0    0.0
1    1.0
dtype: float64

ser3 = Series(range(3), index=[-5, 1, 3])
ser3.iloc[2]

2

frame = DataFrame(np.arange(6).reshape((3, 2)), index=[2, 0, 1])
frame
frame.iloc[0]

	0	1
2	0	1
0	2	3
1	4	5

0    0
1    1
Name: 2, dtype: int32

面板数据

import pandas_datareader.data as web

pdata = pd.Panel(dict((stk, web.get_data_yahoo(stk))
                       for stk in ['AAPL', 'GOOG', 'MSFT', 'DELL']))

pdata

<class 'pandas.core.panel.Panel'>
Dimensions: 4 (items) x 1820 (major_axis) x 6 (minor_axis)
Items axis: AAPL to MSFT
Major_axis axis: 2010-01-04 00:00:00 to 2017-02-27 00:00:00
Minor_axis axis: Open to Adj Close

pdata = pdata.swapaxes('items', 'minor')
pdata['Adj Close'].iloc[:10]

	AAPL	DELL	GOOG	MSFT
Date
2010-01-04	27.727039	14.06528	313.062468	25.555485
2010-01-05	27.774976	14.38450	311.683844	25.563741
2010-01-06	27.333178	14.10397	303.826685	25.406859
2010-01-07	27.282650	14.23940	296.753749	25.142634
2010-01-08	27.464034	14.36516	300.709808	25.316031
2010-01-11	27.221758	14.37483	300.255255	24.994007
2010-01-12	26.912110	14.56830	294.945572	24.828866
2010-01-13	27.291720	14.57797	293.252243	25.060064
2010-01-14	27.133657	14.22005	294.630868	25.563741
2010-01-15	26.680198	13.92985	289.710772	25.481172

pdata.ix[:, '6/1/2012', :]

	Open	High	Low	Close	Volume	Adj Close
AAPL	569.159996	572.650009	560.520012	560.989983	130246900.0	72.681610
DELL	12.150000	12.300000	12.045000	12.070000	19397600.0	11.675920
GOOG	571.790972	572.650996	568.350996	570.981000	6138700.0	285.205295
MSFT	28.760000	28.959999	28.440001	28.450001	56634300.0	24.942239

pdata.ix['Adj Close', '5/22/2012':, :].iloc[:10]

	AAPL	DELL	GOOG	MSFT
Date
2012-05-22	72.160786	14.58765	300.100412	26.090721
2012-05-23	73.921494	12.08221	304.426106	25.520864
2012-05-24	73.242607	12.04351	301.528978	25.485795
2012-05-25	72.850038	12.05319	295.470050	25.477028
2012-05-28	NaN	12.05319	NaN	NaN
2012-05-29	74.143041	12.24666	296.873645	25.915380
2012-05-30	75.037005	12.14992	293.821674	25.722505
2012-05-31	74.850442	11.92743	290.140354	25.591000
2012-06-01	72.681610	11.67592	285.205295	24.942239
2012-06-04	73.109156	11.60821	289.006480	25.029908

stacked = pdata.ix[:, '5/30/2012':, :].to_frame()
stacked

		Open	High	Low	Close	Volume	Adj Close
Date	minor
2012-05-30	AAPL	569.199997	579.989990	566.559990	579.169998	132357400.0	75.037005
	DELL	12.590000	12.700000	12.460000	12.560000	19787800.0	12.149920
	GOOG	588.161028	591.901014	583.530999	588.230992	3827600.0	293.821674
	MSFT	29.350000	29.480000	29.120001	29.340000	41585500.0	25.722505
2012-05-31	AAPL	580.740021	581.499985	571.460022	577.730019	122918600.0	74.850442
	DELL	12.530000	12.540000	12.330000	12.330000	19955600.0	11.927430
	GOOG	588.720982	590.001032	579.001013	580.860990	5958800.0	290.140354
	MSFT	29.299999	29.420000	28.940001	29.190001	39134000.0	25.591000
2012-06-01	AAPL	569.159996	572.650009	560.520012	560.989983	130246900.0	72.681610
	DELL	12.150000	12.300000	12.045000	12.070000	19397600.0	11.675920
	GOOG	571.790972	572.650996	568.350996	570.981000	6138700.0	285.205295
	MSFT	28.760000	28.959999	28.440001	28.450001	56634300.0	24.942239
2012-06-04	AAPL	561.500008	567.499985	548.499977	564.289978	139248900.0	73.109156
	DELL	12.110000	12.112500	11.800000	12.000000	17015700.0	11.608210
	GOOG	570.220958	580.491016	570.011006	578.590973	4883500.0	289.006480
	MSFT	28.620001	28.780001	28.320000	28.549999	47926300.0	25.029908
2012-06-05	AAPL	561.269989	566.470001	558.330002	562.830025	97053600.0	72.920005
	DELL	11.950000	12.240000	11.950000	12.160000	15620900.0	11.762980
	GOOG	575.451008	578.131003	566.470986	570.410999	4697200.0	284.920579
	MSFT	28.510000	28.750000	28.389999	28.510000	45715400.0	24.994841
2012-06-06	AAPL	567.770004	573.849983	565.499992	571.460022	100363900.0	74.038104
	DELL	12.210000	12.280000	12.090000	12.215000	20779900.0	11.816190
	GOOG	576.480979	581.970971	573.611004	580.570966	4207200.0	289.995487
	MSFT	28.879999	29.370001	28.809999	29.350000	46860500.0	25.731273
2012-06-07	AAPL	577.290009	577.320023	570.500000	571.720001	94941700.0	74.071787
	DELL	12.320000	12.410000	12.120000	12.130000	20074000.0	11.733960
	GOOG	587.601014	587.891038	577.251006	578.230986	3530100.0	288.826666
	MSFT	29.639999	29.700001	29.170000	29.230000	37792800.0	25.626067
2012-06-08	AAPL	571.599998	580.580017	568.999992	580.319984	86879100.0	75.185997
	DELL	12.130000	12.225000	12.020000	12.120000	18155600.0	11.724290
…	…	…	…	…	…	…	…
2017-02-13	AAPL	133.080002	133.820007	132.750000	133.289993	23035400.0	133.289993
	GOOG	816.000000	820.958984	815.489990	819.239990	1198100.0	819.239990
	MSFT	64.239998	64.860001	64.129997	64.720001	22920100.0	64.330000
2017-02-14	AAPL	133.470001	135.089996	133.250000	135.020004	32815500.0	135.020004
	GOOG	819.000000	823.000000	816.000000	820.450012	1053600.0	820.450012
	MSFT	64.410004	64.720001	64.019997	64.570000	23065900.0	64.570000
2017-02-15	AAPL	135.520004	136.270004	134.619995	135.509995	35501600.0	135.509995
	GOOG	819.359985	823.000000	818.469971	818.979980	1304000.0	818.979980
	MSFT	64.500000	64.570000	64.160004	64.529999	16917000.0	64.529999
2017-02-16	AAPL	135.669998	135.899994	134.839996	135.350006	22118000.0	135.350006
	GOOG	819.929993	824.400024	818.979980	824.159973	1281700.0	824.159973
	MSFT	64.739998	65.239998	64.440002	64.519997	20524700.0	64.519997
2017-02-17	AAPL	135.100006	135.830002	135.100006	135.720001	22084500.0	135.720001
	GOOG	823.020020	828.070007	821.655029	828.070007	1597800.0	828.070007
	MSFT	64.470001	64.690002	64.300003	64.620003	21234600.0	64.620003
2017-02-21	AAPL	136.229996	136.750000	135.979996	136.699997	24265100.0	136.699997
	GOOG	828.659973	833.450012	828.349976	831.659973	1247700.0	831.659973
	MSFT	64.610001	64.949997	64.449997	64.489998	19384900.0	64.489998
2017-02-22	AAPL	136.429993	137.119995	136.110001	137.110001	20745300.0	137.110001
	GOOG	828.659973	833.250000	828.640015	830.760010	982900.0	830.760010
	MSFT	64.330002	64.389999	64.050003	64.360001	19259700.0	64.360001
2017-02-23	AAPL	137.380005	137.479996	136.300003	136.529999	20704100.0	136.529999
	GOOG	830.119995	832.460022	822.880005	831.330017	1470100.0	831.330017
	MSFT	64.419998	64.730003	64.190002	64.620003	20235200.0	64.620003
2017-02-24	AAPL	135.910004	136.660004	135.279999	136.660004	21690900.0	136.660004
	GOOG	827.729980	829.000000	824.200012	828.640015	1386600.0	828.640015
	MSFT	64.529999	64.800003	64.139999	64.620003	21705200.0	64.620003
2017-02-27	AAPL	137.139999	137.440002	136.279999	136.929993	20196400.0	136.929993
	GOOG	824.549988	830.500000	824.000000	829.280029	1099500.0	829.280029
	MSFT	64.540001	64.540001	64.050003	64.230003	15850400.0	64.230003

3952 rows × 6 columns

stacked.to_panel()

<class 'pandas.core.panel.Panel'>
Dimensions: 6 (items) x 1207 (major_axis) x 4 (minor_axis)
Items axis: Open to Adj Close
Major_axis axis: 2012-05-30 00:00:00 to 2017-02-27 00:00:00
Minor_axis axis: AAPL to MSFT