pandas基本使用的练习笔记

1. 数据索引

import pandas as pd
left = pd.DataFrame( {
     'key':['k1', 'k2', 'k3', 'k4'],
                      'name':['n1', 'n2', 'n3', 'n4'],
                      'id':['id1', 'id2', 'id3', 'id4']
                     }
                   ) # 数组构成的字典

right = pd.DataFrame( {
     'key':['k221', 'k222', 'k223', 'k224'],
                      'name':['n221', 'n222', 'n223', 'n224'],
                      'id':['id221', 'id222', 'id223', 'id224']
                     }
                   )

print(left)
print(right)

    id key name
0  id1  k1   n1
1  id2  k2   n2
2  id3  k3   n3
3  id4  k4   n4
      id   key  name
0  id221  k221  n221
1  id222  k222  n222
2  id223  k223  n223
3  id224  k224  n224

A 选取行

# 1. 用数值取
left.loc[2]

id      id3
key      k3
name     n3
Name: 2, dtype: object

# 2. 用列表取
left.loc[ [1,2,3]] # 注意外层 【】

	id	key	name
1	id2	k2	n2
2	id3	k3	n3
3	id4	k4	n4

#3. 用切片
left.loc[ 1:2 ]

	id	key	name
1	id2	k2	n2
2	id3	k3	n3

#4. 用bool做索引，先要得到bool series
mask = left['name'] > 'n1'
mask

0    False
1     True
2     True
3     True
Name: name, dtype: bool

#4.1 使用 bool series
left.loc[ mask]

	id	key	name
1	id2	k2	n2
2	id3	k3	n3
3	id4	k4	n4

# 5. iloc
left.iloc[3]

id      id4
key      k4
name     n4
Name: 3, dtype: object

# 5.1 iloc
left.iloc[ [0,1]]

	id	key	name
0	id1	k1	n1
1	id2	k2	n2

# 5.2 iloc && 切片
left.iloc[ 0:2]

	id	key	name
0	id1	k1	n1
1	id2	k2	n2

left.loc[ 0:2] # 注意区别，左闭有闭

	id	key	name
0	id1	k1	n1
1	id2	k2	n2
2	id3	k3	n3

left.set_index('key')

	id	name
key
k1	id1	n1
k2	id2	n2
k3	id3	n3
k4	id4	n4

left.set_index('name')

	id	key
name
n1	id1	k1
n2	id2	k2
n3	id3	k3
n4	id4	k4

left.loc[ 0:2]

	id	key	name
0	id1	k1	n1
1	id2	k2	n2
2	id3	k3	n3

left_reset_index = left.set_index("key")
left_reset_index

	id	name
key
k1	id1	n1
k2	id2	n2
k3	id3	n3
k4	id4	n4

left_reset_index.loc[ ['k1']]

	id	name
key
k1	id1	n1

left_reset_index.iloc[ [0,2]]

	id	name
key
k1	id1	n1
k3	id3	n3

B. 选取列

# 1. 列名
left[ 'name']

0    n1
1    n2
2    n3
3    n4
Name: name, dtype: object

# 2. 列表
left[ ['name', 'id']]

	name	id
0	n1	id1
1	n2	id2
2	n3	id3
3	n4	id4

# 3. 切片
left[ 0:2]  # 变成行了

	id	key	name
0	id1	k1	n1
1	id2	k2	n2

# 4. bool
left[ left['name'] > 'n2']

	id	key	name
2	id3	k3	n3
3	id4	k4	n4

C. 选取行和列

#1. 位置
left
left.loc[0, "key"]

'k1'

# 2. 列表
left.loc[ [1,2], ['name', 'id']]

	name	id
1	n2	id2
2	n3	id3

# 3. 切片
left.loc[ 0:2, :]

	id	key	name
0	id1	k1	n1
1	id2	k2	n2
2	id3	k3	n3

# 4.iloc
left.iloc[ [0,1],[0,2]]

	id	name
0	id1	n1
1	id2	n2

left.iloc[ [0,1],[0,1]]

	id	key
0	id1	k1
1	id2	k2

基本操作

# 得到列名们
left.columns 

Index(['id', 'key', 'name'], dtype='object')

#修改列名
left.columns = ["xuhao", 'guanjianzhi', 'mingzi'] 
left

	xuhao	guanjianzhi	mingzi
0	id1	k1	n1
1	id2	k2	n2
2	id3	k3	n3
3	id4	k4	n4

# 获取的简单方式 。语法
left.xuhao

0    id1
1    id2
2    id3
3    id4
Name: xuhao, dtype: object

# 重命名列
left.rename(columns={
     "xuhao":"ID", "guanjianzhi":"KEY", "mingzi":"NAME"})

	ID	KEY	NAME
0	id1	k1	n1
1	id2	k2	n2
2	id3	k3	n3
3	id4	k4	n4

left

	xuhao	guanjianzhi	mingzi
0	id1	k1	n1
1	id2	k2	n2
2	id3	k3	n3
3	id4	k4	n4

# 就地修改 inplace 参数
left.rename( columns={
     "xuhao":"ID", "guanjianzhi":"KEY", "mingzi":"NAME"}, inplace=True)
left

	ID	KEY	NAME
0	id1	k1	n1
1	id2	k2	n2
2	id3	k3	n3
3	id4	k4	n4

# 修改元素值 字典的字典，外层字典用来指定改哪一列
left.replace( {
     "ID":{
     'id1':"ID1", 'id2':'ID2'}}, inplace=True);
left

	ID	KEY	NAME
0	ID1	k1	n1
1	ID2	k2	n2
2	id3	k3	n3
3	id4	k4	n4

基本的排序和增删

# 有多少种取值可能
left.ID.unique()

array(['ID1', 'ID2', 'id3', 'id4'], dtype=object)

# 每一种取值的数量
left.ID.value_counts()

ID1    1
id3    1
ID2    1
id4    1
Name: ID, dtype: int64

# 排序
left.sort_values('ID')

	ID	KEY	NAME
0	ID1	k1	n1
1	ID2	k2	n2
2	id3	k3	n3
3	id4	k4	n4

# 多个key 排序
left.sort_values(['ID', 'NAME'], ascending=[True, False])

	ID	KEY	NAME
0	ID1	k1	n1
1	ID2	k2	n2
2	id3	k3	n3
3	id4	k4	n4

left.sort_values( ['ID', "NAME"], ascending=[False, False])

	ID	KEY	NAME
3	id4	k4	n4
2	id3	k3	n3
1	ID2	k2	n2
0	ID1	k1	n1

left.replace( {
     "ID":{
     'id3':'ID3', 'ID2':'id2'}}, inplace=True)
left

	ID	KEY	NAME
0	ID1	k1	n1
1	id2	k2	n2
2	ID3	k3	n3
3	id4	k4	n4

left.sort_values(['ID'], ascending=[True])

	ID	KEY	NAME
0	ID1	k1	n1
2	ID3	k3	n3
1	id2	k2	n2
3	id4	k4	n4

left['mailema'] = 1
left

	ID	KEY	NAME	mailema
0	ID1	k1	n1	1
1	id2	k2	n2	1
2	ID3	k3	n3	1
3	id4	k4	n4	1

left['rongyuceshilie'] = 1
left

	ID	KEY	NAME	mailema	rongyuceshilie
0	ID1	k1	n1	1	1
1	id2	k2	n2	1	1
2	ID3	k3	n3	1	1
3	id4	k4	n4	1	1

# 删除列
del left['rongyuceshilie']
left

	ID	KEY	NAME	mailema
0	ID1	k1	n1	1
1	id2	k2	n2	1
2	ID3	k3	n3	1
3	id4	k4	n4	1

# 删除行
ret = left.drop(labels=2)
print(left)
print(ret)

    ID KEY NAME  mailema
0  ID1  k1   n1        1
1  id2  k2   n2        1
2  ID3  k3   n3        1
3  id4  k4   n4        1
    ID KEY NAME  mailema
0  ID1  k1   n1        1
1  id2  k2   n2        1
3  id4  k4   n4        1

# 修改一列的数值 map 结合字典
left.NAME
ret = left.NAME.map({
     'n1':'NNN1','n2':"N2", 'n3':'NN3', 'n4':"NN4"})
print(ret)
print(left.NAME)

0    NNN1
1      N2
2     NN3
3     NN4
Name: NAME, dtype: object
0    n1
1    n2
2    n3
3    n4
Name: NAME, dtype: object

# map 修改一列 map结合函数
import numpy as np
left.NAME.map("hello my name is {}".format)
test_series = left.NAME
yy = pd.Series(['name_ext', np.nan])
print(yy)
test_series = test_series.append(yy)
test_series = test_series.map("my name is {}".format, na_action='ignore')
print(test_series)

0    name_ext
1         NaN
dtype: object
0          my name is n1
1          my name is n2
2          my name is n3
3          my name is n4
0    my name is name_ext
1                    NaN
dtype: object

from collections import Counter
counter = Counter()
counter['bar'] += 1
counter

Counter({'bar': 1})

left.min()

ID                ID1
KEY                k1
NAME               n1
mailema             1
rongyuceshilie      1
dtype: object

left.max()

ID                id4
KEY                k4
NAME               n4
mailema             1
rongyuceshilie      1
dtype: object

left.sum()

ID                ID1id2ID3id4
KEY                   k1k2k3k4
NAME                  n1n2n3n4
mailema                      4
rongyuceshilie               4
dtype: object

left.cumsum()

	ID	KEY	NAME	mailema	rongyuceshilie
0	ID1	k1	n1	1	1
1	ID1id2	k1k2	n1n2	2	2
2	ID1id2ID3	k1k2k3	n1n2n3	3	3
3	ID1id2ID3id4	k1k2k3k4	n1n2n3n4	4	4

pandas 常用操作矩阵运算

# 最值的位置
df = pd.DataFrame(np.random.random( (5, 10)), columns = list('abcdefghij')) # 简写列名
df

	a	b	c	d	e	f	g	h	i	j
0	0.920187	0.468342	0.251023	0.012371	0.410920	0.823567	0.644462	0.971567	0.363926	0.125066
1	0.388891	0.017500	0.400027	0.907302	0.747884	0.850026	0.871119	0.361581	0.313365	0.517795
2	0.858049	0.818444	0.429782	0.953707	0.362399	0.597594	0.583865	0.363218	0.671535	0.887737
3	0.981348	0.884835	0.333632	0.930583	0.990436	0.264681	0.349709	0.938117	0.729409	0.371021
4	0.409097	0.270208	0.093192	0.985581	0.389709	0.891675	0.551112	0.897221	0.573714	0.854950

df.idxmax()

a    3
b    3
c    2
d    4
e    3
f    4
g    1
h    0
i    3
j    2
dtype: int64

df*10

	a	b	c	d	e	f	g	h	i	j
0	9.201869	4.683421	2.510227	0.123709	4.109205	8.235670	6.444622	9.715667	3.639265	1.250659
1	3.888908	0.174995	4.000269	9.073024	7.478841	8.500256	8.711186	3.615811	3.133648	5.177951
2	8.580488	8.184435	4.297818	9.537071	3.623993	5.975941	5.838652	3.632183	6.715348	8.877372
3	9.813483	8.848354	3.336317	9.305834	9.904355	2.646808	3.497094	9.381173	7.294089	3.710212
4	4.090968	2.702080	0.931915	9.855806	3.897086	8.916751	5.511120	8.972206	5.737144	8.549502

df*df

	a	b	c	d	e	f	g	h	i	j
0	0.846744	0.219344	0.063012	0.000153	0.168856	0.678263	0.415332	0.943942	0.132442	0.015641
1	0.151236	0.000306	0.160022	0.823198	0.559331	0.722544	0.758848	0.130741	0.098197	0.268112
2	0.736248	0.669850	0.184712	0.909557	0.131333	0.357119	0.340899	0.131928	0.450959	0.788077
3	0.963045	0.782934	0.111310	0.865985	0.980962	0.070056	0.122297	0.880064	0.532037	0.137657
4	0.167360	0.073012	0.008685	0.971369	0.151873	0.795084	0.303724	0.805005	0.329148	0.730940

df.dot(df.T)

	0	1	2	3	4
0	3.483729	2.576565	3.018221	3.486335	2.975670
1	2.576565	3.672533	3.474288	3.405148	3.571698
2	3.018221	3.474288	4.700682	4.478348	4.018175
3	3.486335	3.405148	4.478348	5.446347	3.980907
4	2.975670	3.571698	4.018175	3.980907	4.336201

df.T.dot(df)

	a	b	c	d	e	f	g	h	i	j
a	2.864633	2.118905	1.120862	2.498977	2.111316	2.225693	2.001426	2.633968	1.983464	1.792030
b	2.118905	1.745446	0.796707	1.891951	1.483817	1.364820	1.253284	1.831142	1.525968	1.353505
c	1.120862	0.796707	0.527741	1.178257	0.924835	0.975005	0.929213	0.941231	0.802141	0.823518
d	2.498977	1.891951	1.178257	3.570262	2.335035	2.476474	2.223776	2.443766	2.173484	2.505874
e	2.111316	1.483817	0.924835	2.335035	1.992355	1.800352	1.689049	2.080087	1.573283	1.461012
f	2.225693	1.364820	0.975005	2.476474	1.800352	2.623065	2.204120	2.372892	1.672019	1.934186
g	2.001426	1.253284	0.929213	2.223776	1.689049	2.204120	1.941099	1.975726	1.470863	1.650903
h	2.633968	1.831142	0.941231	2.443766	2.080087	2.372892	1.975726	2.891679	1.909819	1.746318
i	1.983464	1.525968	0.802141	2.173484	1.573283	1.672019	1.470863	1.909819	1.542785	1.565043
j	1.792030	1.353505	0.823518	2.505874	1.461012	1.934186	1.650903	1.746318	1.565043	1.940427