python——pandas库之DataFrame数据结构基础

DataFrame简介

dataframe表示的是矩阵的数据表，每一列可以是不同的值类型，可以看作一个共享相同索引的Series字典。在dataframe中，数据被存储为一个以上的二维块

1.利用等长度列表或numpy数组的字典来形成dataframe

import pandas as pd
import numpy as np

data = {
     'state' : ['ohio', 'ohio', 'ohio', 'Nevada', 'Nevada', 'Nevada'],
        'year' : [2000, 2001, 2002, 2001, 2002, 2003],
        'pop' : [1.5, 1.7, 3.6, 2.4, 2.9, 3.2]}
frame = pd.DataFrame(data)
frame

	state	year	pop
0	ohio	2000	1.5
1	ohio	2001	1.7
2	ohio	2002	3.6
3	Nevada	2001	2.4
4	Nevada	2002	2.9
5	Nevada	2003	3.2

2.对于大型DataFrame，head()方法将会只选出头部的五行

frame.head()

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

3.如果指定了列的顺序，DataFrame的列将会按照指定顺序排列

pd.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
5	2003	Nevada	3.2

# 如果传入的列不再字典中，将会在结果中出现缺失值
frame2 = pd.DataFrame(data, columns = ['year', 'state', 'pop', 'debt'], index = ['one', 'two', 'three', 'four', 'five', 'six'])
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
six	2003	Nevada	3.2	NaN

frame2.columns

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

4.DataFrame中的一列，可以按照字典型标记或属性那样检索为Series

frame2['state']

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

frame2.year

one      2000
two      2001
three    2002
four     2001
five     2002
six      2003
Name: year, dtype: int64

5.通过位置或属性loc进行行的选取

frame2.loc['three']

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

6.修改列的值

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
six	2003	Nevada	3.2	16.5

frame2['ddebt'] = np.arange(6.)
frame2

	year	state	pop	debt	ddebt
one	2000	ohio	1.5	16.5	0.0
two	2001	ohio	1.7	16.5	1.0
three	2002	ohio	3.6	16.5	2.0
four	2001	Nevada	2.4	16.5	3.0
five	2002	Nevada	2.9	16.5	4.0
six	2003	Nevada	3.2	16.5	5.0

7.当你将列表或数组赋值给一个列时，值的长度必须和DataFrame的长度相匹配，如果你将Series赋值给一列，Series的索引将会按照DataFrame的索引重新排序，并在孔雀的地方填充空缺值

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

	year	state	pop	debt	ddebt
one	2000	ohio	1.5	NaN	0.0
two	2001	ohio	1.7	-1.2	1.0
three	2002	ohio	3.6	NaN	2.0
four	2001	Nevada	2.4	-1.5	3.0
five	2002	Nevada	2.9	-1.7	4.0
six	2003	Nevada	3.2	NaN	5.0

如果被赋值的列并不存在，则会生成一个新的列，del关键字可以删除列

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

	year	state	pop	debt	ddebt	eastern
one	2000	ohio	1.5	NaN	0.0	True
two	2001	ohio	1.7	-1.2	1.0	True
three	2002	ohio	3.6	NaN	2.0	True
four	2001	Nevada	2.4	-1.5	3.0	False
five	2002	Nevada	2.9	-1.7	4.0	False
six	2003	Nevada	3.2	NaN	5.0	False

del frame2['eastern']
frame2

	year	state	pop	debt	ddebt
one	2000	ohio	1.5	NaN	0.0
two	2001	ohio	1.7	-1.2	1.0
three	2002	ohio	3.6	NaN	2.0
four	2001	Nevada	2.4	-1.5	3.0
five	2002	Nevada	2.9	-1.7	4.0
six	2003	Nevada	3.2	NaN	5.0

8.字典的嵌套字典

如果嵌套字典被赋值给dataframe，pandas会将字典的键作为列，将内部字典的键作为行索引

pop = {
     'Nevada':{
     2001: 2.4, 2002: 2.9}, 'Ohio' : {
     2000:1.5, 2001:1.7, 2002:3.6}}
frame3 = pd.DataFrame(pop)
frame3

	Nevada	Ohio
2001	2.4	1.7
2002	2.9	3.6
2000	NaN	1.5

# 显示指明索引的顺序
pd.DataFrame(pop, index = [2001, 2002, 2003])

	Nevada	Ohio
2001	2.4	1.7
2002	2.9	3.6
2003	NaN	NaN

包含Series的字典也可以用于构造DataFrame

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

	Ohio	Nevada
2001	1.7	2.4
2002	3.6	2.9

DataFrame的索引和列拥有name属性

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

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

values属性得到dataframe中的值

frame3.values

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

frame2.values

array([[2000, 'ohio', 1.5, nan, 0.0],
       [2001, 'ohio', 1.7, -1.2, 1.0],
       [2002, 'ohio', 3.6, nan, 2.0],
       [2001, 'Nevada', 2.4, -1.5, 3.0],
       [2002, 'Nevada', 2.9, -1.7, 4.0],
       [2003, 'Nevada', 3.2, nan, 5.0]], dtype=object)