Python Pandas-使用部分匹配合并两个数据帧

有了下图所示的数据帧，我想先在

['A'、'B'、'C']

和

['X'、'Y'、'Z']

上合并，然后逐渐寻找一个少一列的匹配项，即

['A'、'B']

和

['X'、'Y']

然后

['A']

和

['X']

，而不复制结果行，在下面的示例中，

a，y，y，v3

被省略，因为

a，d，d

已经匹配

到目前为止，我的代码在所有3列上都匹配：

df1 = pd.DataFrame({"A":['a','b','c'],"B":['d','e','f'],"C":['d','e','f']})
df2 = pd.DataFrame({"X":['a','b','a','c'],"Y":['d','e','y','z'],"Z":['d','x','y','z'],"V":['v1','v2','v3','v4']})

merged = pd.merge(df1,df2,left_on=['A','B','C'],right_on=['X','Y','Z'], how='left')
merged = merged.drop_duplicates(['A','B','C'])
merged.head()

---

我怎样才能实现我的目标

更新：预期输出

这个怎么样：

matches = [['A', 'B', 'C'], ['X', 'Y', 'Z']]
df = df1.copy()
for k in range(len(matches[0])):

    #Get your left/right keys right at each iteration :
    left, right = matches
    left = left if k==0 else left[:-k]
    right = right if k==0 else right[:-k]

    #Make sure columns from df2 exist in df
    for col in df2.columns.tolist():
        try:
            df[col]
        except Exception:
            df[col] = np.nan

    #Merge dataframes
    df = df.merge(df2, left_on=left, right_on=right, how='left')

    #Find which row of df's "left" columns (previously initialised) are empty
    ix_left_part = np.all([df[x + "_x"].isnull() for x in right], axis=0)

    #Find which row of df's "right" columns are not empty
    ix_right_part = np.all([df[x + "_y"].notnull() for x in right], axis=0)

    #Combine both to get indexes
    ix = df[ix_left_part & ix_right_part].index

    #Complete values on "left" with those from "right"
    for x in df2.columns.tolist():
        df.loc[ix, x+"_x"] = df.loc[ix, x+'_y']

    #Drop values from "right"
    df.drop([x+"_y" for x  in df2.columns.tolist()], axis=1, inplace=True)

    #Rename "left" columns to stick with original names from df2
    df.rename({x+"_x":x for x  in df2.columns.tolist()}, axis=1, inplace=True)

#drop eventual duplicates
df.drop_duplicates(keep="first", inplace=True)
print(df)

---

编辑

我纠正了这个循环；这对内存来说应该更容易：

import pandas as pd
import numpy as np

df1 = pd.DataFrame({"A":['a','b','c'],"B":['d','e','f'],"C":['d','e','f']})
df2 = pd.DataFrame({"X":['a','b','a','c'],"Y":['d','e','y','z'],"Z":['d','x','y','z'],"V":['v1','v2','v3','v4']})

matches = [['A', 'B', 'C'], ['X', 'Y', 'Z']]
df = df1.copy()

#Make sure columns of df2 exist in df
for col in df2.columns.tolist():
    df[col] = np.nan

for k in range(len(matches[0])):

    #Get your left/right keys right at each iteration :
    left, right = matches
    left = left if k==0 else left[:-k]
    right = right if k==0 else right[:-k]
    
    #recreate dataframe of (potential) usable datas in df2:
    ix = df[df.V.isnull()].index
    temp = (
            df.loc[ix, left]
            .rename(dict(zip(left, right)), axis=1)
            )
    
    temp=temp.merge(df2, on=right, how="inner")
    
    #Merge dataframes
    df = df.merge(temp, left_on=left, right_on=right, how='left')
    
    
    #Combine both to get indexes
    ix = df[(df['V_x'].isnull()) & (df['V_y'].notnull())].index
    

    #Complete values on "left" with those from "right"
    cols_left = [x+'_x' for x in df2.columns.tolist()]
    cols_right = [x+'_y' for x in df2.columns.tolist()]    
    df.loc[ix, cols_left] = df.loc[ix, cols_right].values.tolist()
        
    #Drop values from "right"
    df.drop(cols_right, axis=1, inplace=True)
    
    #Rename "left" columns to stick with original names from df2
    rename = {x+"_x":x for x  in df2.columns.tolist()}
    df.rename(rename, axis=1, inplace=True)

print(df)

---

一个想法与多个

合并

在循环中，第二个

数据帧

应该避免在最终的

数据帧

中重复行：

from functools import reduce

dfs = []
L = [['A', 'B', 'C'], ['X', 'Y', 'Z']]

for i in range(len(L[0]), 0, -1):
    df22 = df2.drop_duplicates(L[1][:i])
    df = pd.merge(df1,df22,left_on=L[0][:i],right_on=L[1][:i], how='left')
    dfs.append(df)

df = reduce(lambda l,r: pd.DataFrame.fillna(l,r), dfs)
print (df)
   A  B  C  X  Y  Z   V
0  a  d  d  a  d  d  v1
1  b  e  e  b  e  x  v2
2  c  f  f  c  z  z  v4

工作方式如下：

merged1 = pd.merge(df1,df2.drop_duplicates(['X','Y','Z']),left_on=['A','B','C'],right_on=['X','Y','Z'], how='left')
merged2 = pd.merge(df1,df2.drop_duplicates(['X','Y']),left_on=['A','B'],right_on=['X','Y'], how='left')
merged3 = pd.merge(df1,df2.drop_duplicates('X'),left_on=['A'],right_on=['X'], how='left')

df = merged1.fillna(merged2).fillna(merged3)
print (df)
   A  B  C  X  Y  Z   V
0  a  d  d  a  d  d  v1
1  b  e  e  b  e  x  v2
2  c  f  f  c  z  z  v4

---

a，y，y，v3

已被忽略，因为您已经有一行与整个3列匹配？你能不能也加上预期的输出？没错，我会加上预期的输出。我想我把“ix_left_part”搞砸了，你不应该在最后得到重复的。。。我想我现在知道它是什么了，但是纠正它将取决于是否有一个“target”列（在您的实际df2上是“V”的意思）。有一个目标，V列。我将测试代码，让您知道您的代码在我提供的虚拟数据上工作，这是值得称赞的…但是，当我在实际数据上运行它时，它会消耗大量内存并使我的环境崩溃：（数据帧的形状是什么？df1是（10000，3）而df2是（137503，3），在我的实际数据中只有两列需要匹配，但我正在寻找一个通用的答案，就像你提供的答案一样。超级！工作得像个charmcan你可以看看这里：？