顯示 3.2 選擇 列數 行合並 重復 原來 result value

1，pd.concat：拼接
1.1，axis
df1 = pd.DataFrame(np.ones((3,4))*0, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
df2 = pd.DataFrame(np.ones((3,4))*1, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
df3 = pd.DataFrame(np.ones((3,4))*2, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
a b c d
0 0.0 0.0 0.0 0.0
1 0.0 0.0 0.0 0.0
2 0.0 0.0 0.0 0.0
a b c d
0 1.0 1.0 1.0 1.0
1 1.0 1.0 1.0 1.0
2 1.0 1.0 1.0 1.0
a b c d
0 2.0 2.0 2.0 2.0
1 2.0 2.0 2.0 2.0
2 2.0 2.0 2.0 2.0 result = pd.concat([df1, df2, df3], axis = 0) # 0表示豎向的合並
a b c d
0 0.0 0.0 0.0 0.0
1 0.0 0.0 0.0 0.0
2 0.0 0.0 0.0 0.0
0 1.0 1.0 1.0 1.0
1 1.0 1.0 1.0 1.0
2 1.0 1.0 1.0 1.0
0 2.0 2.0 2.0 2.0
1 2.0 2.0 2.0 2.0
2 2.0 2.0 2.0 2.0 result=pd.concat([df1,df2,df3], axis=1) # 1表示橫向的合並
a b c d a b c d a b c d
0 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0
1 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0
2 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0 1.2，ignore_index
df1 = pd.DataFrame(np.ones((3,4))*0, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
df2 = pd.DataFrame(np.ones((3,4))*1, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
df3 = pd.DataFrame(np.ones((3,4))*2, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
result = pd.concat([df1, df2, df3], axis = 0, ignore_index=True) # ignore_index=True表示忽略原來index(axis=0)/column(axis=1)
a b c d
0 0.0 0.0 0.0 0.0
1 0.0 0.0 0.0 0.0
2 0.0 0.0 0.0 0.0
3 1.0 1.0 1.0 1.0
4 1.0 1.0 1.0 1.0
5 1.0 1.0 1.0 1.0
6 2.0 2.0 2.0 2.0
7 2.0 2.0 2.0 2.0
8 2.0 2.0 2.0 2.0 result = pd.concat([df1, df2, df3], axis = 0, ignore_index=True) # ignore_index=True表示忽略原來index
0 1 2 3 4 5 6 7 8 9 10 11
0 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0
1 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0
2 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0 1.3，join
df1 = pd.DataFrame(np.ones((3,4))*0, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
df2 = pd.DataFrame(np.ones((3,4))*1, columns = [‘b‘, ‘c‘, ‘d‘, ‘e‘])
result=pd.concat([df1,df2], axis=0, join=‘inner‘, ignore_index=True) # join=‘inner‘表示只保留column名相同的部分
b c d
0 0.0 0.0 0.0
1 0.0 0.0 0.0
2 0.0 0.0 0.0
3 1.0 1.0 1.0
4 1.0 1.0 1.0
5 1.0 1.0 1.0 1.4, join_axes
df1 = pd.DataFrame(np.ones((3,4))*0, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘], index=[1, 2, 3])
df2 = pd.DataFrame(np.ones((3,4))*1, columns = [‘b‘, ‘c‘, ‘d‘, ‘e‘], index=[2, 3, 4]) result=pd.concat([df1,df2], axis=1)
a b c d b c d e
1 0.0 0.0 0.0 0.0 NaN NaN NaN NaN
2 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0
3 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0
4 NaN NaN NaN NaN 1.0 1.0 1.0 1.0 result=pd.concat([df1,df2], axis=1, join_axes=[df1.index]) # join_axes=[df1.index]表示按照df1的index進行合並
a b c d b c d e
1 0.0 0.0 0.0 0.0 NaN NaN NaN NaN
2 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0
3 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2，pd.append()
2.1，append
pd.append只有豎向往下加數據，沒有axis屬性
df1 = pd.DataFrame(np.ones((3,4))*0, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘], index=[1, 2, 3])
df2 = pd.DataFrame(np.ones((3,4))*1, columns = [‘b‘, ‘c‘, ‘d‘, ‘e‘], index=[2, 3, 4])
df3 = pd.DataFrame(np.ones((3,4))*2, columns = [‘b‘, ‘c‘, ‘d‘, ‘e‘], index=[2, 3, 4]) df1 = df1.append(df2)
a b c d e
1 0.0 0.0 0.0 0.0 NaN
2 0.0 0.0 0.0 0.0 NaN
3 0.0 0.0 0.0 0.0 NaN
2 NaN 1.0 1.0 1.0 1.0
3 NaN 1.0 1.0 1.0 1.0
4 NaN 1.0 1.0 1.0 1.0 df1 = df1.append([df2, df3], ignore_index=True) # 可以一次加多個df，也可以忽略index
a b c d e
0 0.0 0.0 0.0 0.0 NaN
1 0.0 0.0 0.0 0.0 NaN
2 0.0 0.0 0.0 0.0 NaN
3 NaN 1.0 1.0 1.0 1.0
4 NaN 1.0 1.0 1.0 1.0
5 NaN 1.0 1.0 1.0 1.0
6 NaN 2.0 2.0 2.0 2.0
7 NaN 2.0 2.0 2.0 2.0
8 NaN 2.0 2.0 2.0 2.0 s1=pd.Series([1,2,3,4],index=[‘a‘,‘b‘,‘c‘,‘d‘])
df1 = df1.append(s1, ignore_index=True) # 可以加series，ignore_index=True必須加上，s1個數必須和df1的列數相同
a b c d
0 0.0 0.0 0.0 0.0 # 註意ignore_index = True以後，df1的index變成從0開始了！！！
1 0.0 0.0 0.0 0.0
2 0.0 0.0 0.0 0.0
3 1.0 2.0 3.0 4.0 l1=[1,2,3,4]
df1.loc[4] = l1
a b c d
1 0.0 0.0 0.0 0.0 # df1的index還是按照原來從1開始
2 0.0 0.0 0.0 0.0
3 0.0 0.0 0.0 0.0
4 1.0 2.0 3.0 4.0 2.2，append舉例
df1 = pd.DataFrame({‘B‘: [‘B0‘, ‘B1‘, ‘B2‘, ‘B3‘,‘B4‘],
‘D‘: [‘D0‘, ‘D1‘, ‘D2‘, ‘D3‘,‘D4‘],
‘A‘: [‘A0‘, ‘A1‘, ‘A2‘, ‘A3‘,‘A4‘],
‘E‘: [‘E0‘, ‘E1‘, ‘E2‘, ‘E3‘,‘E4‘]},
index=[0, 1, 2, 3,4],columns=[‘B‘,‘D‘,‘A‘,‘E‘]) df2 = pd.DataFrame({‘F‘: [‘F4‘, ‘F5‘, ‘F6‘, ‘F7‘,‘F8‘],
‘A‘: [‘A4‘, ‘A5‘, ‘A6‘, ‘A7‘,‘A8‘],
‘B‘: [‘B4‘, ‘B5‘, ‘B6‘, ‘B7‘,‘B8‘],
‘C‘: [‘C4‘, ‘C5‘, ‘C6‘, ‘C7‘,‘C8‘]},
index=[5, 9, 6, 7,10]) 輸入df3=df1.append(df2)後，df3變為：
A B C D E F
0 A0 B0 NaN D0 E0 NaN
1 A1 B1 NaN D1 E1 NaN
2 A2 B2 NaN D2 E2 NaN
3 A3 B3 NaN D3 E3 NaN
4 A4 B4 NaN D4 E4 NaN
5 A4 B4 C4 NaN NaN F4
9 A5 B5 C5 NaN NaN F5
6 A6 B6 C6 NaN NaN F6
7 A7 B7 C7 NaN NaN F7
10 A8 B8 C8 NaN NaN F8

列項重新排列了，如果想要列項按照df1的走，即df1的次序，並且如果df1沒有的列，就直接扔掉： df4=pd.concat([df1,df2],axis=0)
df4.reindex(columns=df1.columns)
3，pd.merge # 合並
3.1，1個key
left=pd.DataFrame({‘key‘:[‘K0‘,‘K1‘,‘K2‘,‘K3‘],
‘A‘:[‘A0‘,‘A1‘,‘A2‘,‘A3‘],
‘B‘:[‘B0‘,‘B1‘,‘B2‘,‘B3‘]})
right=pd.DataFrame({‘key‘:[‘K0‘,‘K1‘,‘K2‘,‘K3‘],
‘C‘:[‘C0‘,‘C1‘,‘C2‘,‘C3‘],
‘D‘:[‘D0‘,‘D1‘,‘D2‘,‘D3‘]}) res=pd.merge(left,right)
res=pd.merge(left,right,on=[‘key‘]) # 默認會自動選擇共同的column進行合並，如果沒有共同的column會報錯
A B key C D
0 A0 B0 K0 C0 D0
1 A1 B1 K1 C1 D1
2 A2 B2 K2 C2 D2
3 A3 B3 K3 C3 D3 3.2，多個key，how可取[‘left‘, ‘right‘, ‘inner‘, ‘outer‘]
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‘]})
A B key1 key2
0 A0 B0 K0 K0
1 A1 B1 K0 K1
2 A2 B2 K1 K0
3 A3 B3 K2 K1
C D key1 key2
0 C0 D0 K0 K0
1 C1 D1 K1 K0
2 C2 D2 K1 K0
3 C3 D3 K2 K0 res=pd.merge(left,right,on=[‘key1‘,‘key2‘]) # 需要key1，key2完全相同才合並，left有一行K1/K0，right有兩行K1/K0，則打印2遍
res=pd.merge(left,right,on=[‘key1‘,‘key2‘], how = ‘inner‘) # 默認是how=‘inner‘，即key的value要相同
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
1 A2 B2 K1 K0 C1 D1
2 A2 B2 K1 K0 C2 D2 res=pd.merge(left,right,on=[‘key1‘,‘key2‘], how = ‘outer‘) # 根據key1，key2取值全部合並，如果left或者right沒有的取NaN
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
1 A1 B1 K0 K1 NaN NaN
2 A2 B2 K1 K0 C1 D1
3 A2 B2 K1 K0 C2 D2
4 A3 B3 K2 K1 NaN NaN
5 NaN NaN K2 K0 C3 D3 res=pd.merge(left,right,on=[‘key1‘,‘key2‘], how = ‘left‘) # 根據left的key合並，重復出現了2行K1K0（因為right有2行）
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
1 A1 B1 K0 K1 NaN NaN
2 A2 B2 K1 K0 C1 D1
3 A2 B2 K1 K0 C2 D2
4 A3 B3 K2 K1 NaN NaN res=pd.merge(left,right,on=[‘key1‘,‘key2‘],how=‘right‘) # 根據right的key合並
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
1 A2 B2 K1 K0 C1 D1
2 A2 B2 K1 K0 C2 D2
3 NaN NaN K2 K0 C3 D3 3.3, indicator
默認false，true表示顯示合並方式，是left還是right，還是both

dataframe的合並（append, merge, concat）