DataFrame

文章目录

DataFrame行列名称获取与修改按照行名和列名获取数据按照索引获取数据混合调用添加删除修改合并拼接追加

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行列名称获取与修改

import pandas as pd data = { 'state':['Ohio','Ohio','Ohio','Nevada','Nevada'], 'year':[2000,2001,2002,2001,2002], 'pop':[1.5,1.7,3.6,2.4,2.9] } df = pd.DataFrame(data) #获取行名列名 print(df.index) print(df.columns) #修改行名列名 df.index=[0,1,'a','b','3'] df.columns=['one','two','three'] print(df)

输出： 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 RangeIndex(start=0, stop=5, step=1) Index([‘state’, ‘year’, ‘pop’], dtype=‘object’) one two three 0 Ohio 2000 1.5 1 Ohio 2001 1.7 a Ohio 2002 3.6 b Nevada 2001 2.4 3 Nevada 2002 2.9

按照行名和列名获取数据

#根据行名，列名获取数据 print(df.loc[[1,2]])#没有的补位NaN print(df.loc[:,['one']]) print(df.loc[[1,2],['one']]) #一下是简化写法 print(df[['one','two']])#只选取列 #如果只有一个的话可以用一层括号，返回series不是dataframe print(type(df.loc[1]))#选取一行 print(df['one'])

输出： one two three 1 Ohio 2001.0 1.7 2 NaN NaN NaN one 0 Ohio 1 Ohio a Ohio b Nevada 3 Nevada one 1 Ohio 2 NaN one two 0 Ohio 2000 1 Ohio 2001 a Ohio 2002 b Nevada 2001 3 Nevada 2002 <class ‘pandas.core.series.Series’> 0 Ohio 1 Ohio a Ohio b Nevada 3 Nevada Name: one, dtype: object

按照索引获取数据

#根据索引，获取 print(df.iloc[[1,2]])#没有的补位NaN print(df.iloc[:,[1]])#没有的补位NaN print(df.iloc[[1,2],[0,2]])#没有的补位NaN #一下是一些简化写法 df.iloc[1,2] df.iloc[2]#只有一行或者一列的时候用 df.iloc[:,2]

输出： one two three 1 Ohio 2001 1.7 a Ohio 2002 3.6 two 0 2000 1 2001 a 2002 b 2001 3 2002 one three 1 Ohio 1.7 a Ohio 3.6 0 1.5 1 1.7 a 3.6 b 2.4 3 2.9 Name: three, dtype: float64

混合调用

#混合调用 #print(type(df['one'])) print(df['one'].iloc[1:3])#df['one']返回series print(df.iloc[1:3]['two'])#df.iloc[1:3]返回dataframe

输出： 1 Ohio a Ohio Name: one, dtype: object 1 2001 a 2002 Name: two, dtype: int64

添加删除修改

data = { 'state':['Ohio','Ohio','Ohio','Nevada','Nevada'], 'year':[2000,2001,2002,2001,2002], 'pop':[1.5,1.7,3.6,2.4,2.9] } df = pd.DataFrame(data) print(df) #添加、删除、修改 #添加列 df.loc[:,'four']=[1,2,3,4,5]#没有就添加，有就修改 #df['four']=[1,2,3,4,5]#简写 print(df) #添加行 df.loc['last']=[5,5,5,5]#没有就添加，有就修改 print(df) #删除列 df=df.drop(['pop'],axis = 1)#没有就报错 print(1,df) df = df.drop([3,4],axis = 0)#没有就报错 print(df) #插入列 col_name = df.columns.tolist() col_name.insert(1,'city') col_name.pop() df = df.reindex(columns = col_name)#reindex就是使用新的行列名，并按照之前顺序,之前没有的添加为NaN，之前有现在没有的就删了 print(2,df,sep = '\n') df['city']=[8,8,8,8] print(df) #插入列 ind_name = df.index.tolist() ind_name.insert(1,'charu') df = df.reindex(index = ind_name) print(df) df.loc['charu'] = [9,9,9] print(df)

输出： 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 state year pop four 0 Ohio 2000 1.5 1 1 Ohio 2001 1.7 2 2 Ohio 2002 3.6 3 3 Nevada 2001 2.4 4 4 Nevada 2002 2.9 5 state year pop four 0 Ohio 2000 1.5 1 1 Ohio 2001 1.7 2 2 Ohio 2002 3.6 3 3 Nevada 2001 2.4 4 4 Nevada 2002 2.9 5 last 5 5 5.0 5 1 state year four 0 Ohio 2000 1 1 Ohio 2001 2 2 Ohio 2002 3 3 Nevada 2001 4 4 Nevada 2002 5 last 5 5 5 state year four 0 Ohio 2000 1 1 Ohio 2001 2 2 Ohio 2002 3 last 5 5 5 2 state city year 0 Ohio NaN 2000 1 Ohio NaN 2001 2 Ohio NaN 2002 last 5 NaN 5 state city year 0 Ohio 8 2000 1 Ohio 8 2001 2 Ohio 8 2002 last 5 8 5 state city year 0 Ohio 8.0 2000.0 charu NaN NaN NaN 1 Ohio 8.0 2001.0 2 Ohio 8.0 2002.0 last 5 8.0 5.0 state city year 0 Ohio 8.0 2000.0 charu 9 9.0 9.0 1 Ohio 8.0 2001.0 2 Ohio 8.0 2002.0 last 5 8.0 5.0

合并拼接追加

concat

""" pd.concat(objs, axis=0, join='outer', join_axes=None, ignore_index=False, keys=None, levels=None, names=None, verify_integrity=False, copy=True) """ #合并两个 import numpy as np df1 = pd.DataFrame(np.ones((4, 4))*1, columns=list('DCBA'), index=list('4321')) df2 = pd.DataFrame(np.ones((4, 4))*2, columns=list('FEDC'), index=list('6543')) df3 = pd.DataFrame(np.ones((4, 4))*2, columns=list('FEDC'), index=list('6543')) print(df1) print(df2) print(pd.concat([df1, df2])) print(pd.concat([df1, df2], axis=1)) #默认值：axis=0 #axis=0：竖方向（index）合并，合并方向index作列表相加(可重复)，非合并方向columns取并集 #axis=1：横方向（columns）合并，合并方向columns作列表相加（可重复），非合并方向index取并集 #备注：原df中，取并集的行/列名称不能有重复项，即axis=0时columns不能有重复项，axis=1时index不能有重复项：对于参与和并单个矩阵而言 df1.columns = list('DDBA') print(df1) pd.concat([df1, df2], axis=0)#ValueError: Plan shapes are not aligned

输出： D C B A 4 1.0 1.0 1.0 1.0 3 1.0 1.0 1.0 1.0 2 1.0 1.0 1.0 1.0 1 1.0 1.0 1.0 1.0 F E D C 6 2.0 2.0 2.0 2.0 5 2.0 2.0 2.0 2.0 4 2.0 2.0 2.0 2.0 3 2.0 2.0 2.0 2.0 A B C D E F 4 1.0 1.0 1.0 1.0 NaN NaN 3 1.0 1.0 1.0 1.0 NaN NaN 2 1.0 1.0 1.0 1.0 NaN NaN 1 1.0 1.0 1.0 1.0 NaN NaN 6 NaN NaN 2.0 2.0 2.0 2.0 5 NaN NaN 2.0 2.0 2.0 2.0 4 NaN NaN 2.0 2.0 2.0 2.0 3 NaN NaN 2.0 2.0 2.0 2.0 D C B A F E D C 1 1.0 1.0 1.0 1.0 NaN NaN NaN NaN 2 1.0 1.0 1.0 1.0 NaN NaN NaN NaN 3 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0 4 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0 5 NaN NaN NaN NaN 2.0 2.0 2.0 2.0 6 NaN NaN NaN NaN 2.0 2.0 2.0 2.0 D D B A 4 1.0 1.0 1.0 1.0 3 1.0 1.0 1.0 1.0 2 1.0 1.0 1.0 1.0 1 1.0 1.0 1.0 1.0

merge

""" merge(left, right, how='inner', on=None, left_on=None, right_on=None, left_index=False, right_index=False, sort=False, suffixes=('_x', '_y'), copy=True, indicator=False, validate=None) 基于一列或多列，列名要在两个矩阵中都能找到，进行拼接 """ left = pd.DataFrame({'key1': ['K0', 'K0', 'K1', 'K2'], 'key2': ['K0', 'K1', 'K0', 'K1'],'A': ['A0', 'A1', 'A2', 'A3'], 'B': ['B0', 'B1', 'B2', 'B3']}) right = pd.DataFrame({'key1': ['K0', 'K1', 'K1', 'K2'],'key2': ['K0', 'K0', 'K0', 'K0'],'C': ['C0', 'C1', 'C2', 'C3'],'D': ['D0', 'D1', 'D2', 'D3']}) print(left) print(right) #默认相同列名作为连接键，从第一个里面找 K0 K0在right中有，K0，K1在right中没有，K1 K0在right中有两个，K2 K1没有 """ key1 key2 A B C D 0 K0 K0 A0 B0 C0 D0 1 K1 K0 A2 B2 C1 D1 2 K1 K0 A2 B2 C2 D2 #因为AB只有一次K1 K0所以复制一次 """ print(pd.merge(left, right)) print(pd.merge(left, right, on=['key1', 'key2'])) #以右边矩阵为基础,没有的就NAN print(pd.merge(left, right, how='right', on=['key1', 'key2'])) #求并,保留所有行，没有的补充NAN result = pd.merge(left, right, how='outer', on=['key1', 'key2']) print(result)

输出： key1 key2 A B 0 K0 K0 A0 B0 1 K0 K1 A1 B1 2 K1 K0 A2 B2 3 K2 K1 A3 B3 key1 key2 C D 0 K0 K0 C0 D0 1 K1 K0 C1 D1 2 K1 K0 C2 D2 3 K2 K0 C3 D3 key1 key2 A B C D 0 K0 K0 A0 B0 C0 D0 1 K1 K0 A2 B2 C1 D1 2 K1 K0 A2 B2 C2 D2 key1 key2 A B C D 0 K0 K0 A0 B0 C0 D0 1 K1 K0 A2 B2 C1 D1 2 K1 K0 A2 B2 C2 D2 key1 key2 A B C D 0 K0 K0 A0 B0 C0 D0 1 K1 K0 A2 B2 C1 D1 2 K1 K0 A2 B2 C2 D2 3 K2 K0 NaN NaN C3 D3 key1 key2 A B C D 0 K0 K0 A0 B0 C0 D0 1 K0 K1 A1 B1 NaN NaN 2 K1 K0 A2 B2 C1 D1 3 K1 K0 A2 B2 C2 D2 4 K2 K1 A3 B3 NaN NaN 5 K2 K0 NaN NaN C3 D3

update

""" updata主要是数据更新，尽可能多的更新 """ df = pd.DataFrame({'A': [1, 2, 3], 'B': [400, 500, 600]}) print(df) new_df = pd.DataFrame({'B': [4, 5, 6, 7, 8], 'C': [7, 8, 9, 10, 11]})#根据列名和行名进行匹配 df.update(new_df) print(df) df = pd.DataFrame({'A': [1, 2, 3], 'B': [400, 500, 600]}) df.index= list("ABC") print(df.index) new_df = pd.DataFrame({'B': [4, 5, 6, 7, 8], 'C': [7, 8, 9, 10, 11]})#根据列名和行index进行匹配,而鄙视行名成 df.update(new_df) print(df) #使用series更新 new_column = pd.Series(['d', 'e'], name='B', index=list("AB")) print(new_column) df.update(new_column)#替换中如果新表中数据是NAN则不替换 print(df)

A B 0 1 400 1 2 500 2 3 600 A B 0 1 4.0 1 2 5.0 2 3 6.0 Index([‘A’, ‘B’, ‘C’], dtype=‘object’) A B A 1 400 B 2 500 C 3 600 A d B e Name: B, dtype: object A B A 1 d B 2 e C 3 600

append

""" append """ df = pd.DataFrame([[1, 2], [3, 4]], columns=list('AB')) df2 = pd.DataFrame([[5, 6,7], [7, 8,9],[9,10,11]], columns=list('ABC')) print(df) print(df2) print(df.append(df2))#列为两个列的并 print(df.append(df2,ignore_index=True))

输出： A B 0 1 2 1 3 4 A B C 0 5 6 7 1 7 8 9 2 9 10 11 A B C 0 1 2 NaN 1 3 4 NaN 0 5 6 7.0 1 7 8 9.0 2 9 10 11.0 A B C 0 1 2 NaN 1 3 4 NaN 2 5 6 7.0 3 7 8 9.0 4 9 10 11.0