比较PandaS DataFrames并返回从第一个丢失的行 [英] Compare PandaS DataFrames and return rows that are missing from the first one

问题描述

我有2个数据帧，并且要比较它们，并返回不在第二个数据帧（df2）的第一个数据帧（df1）。我找到了一种比较它们并返回差异的方式，但是无法弄清楚如何只从df1返回缺失的一个。

 ，DataFrame 
 
 df1 = pd.DataFrame（{
City：[Chicago，San Franciso，Boston ]，
State：[Illinois，California，Massachusett]}）
 
 df2 = pd.DataFrame（{
City 芝加哥，Mmmmiami，达拉斯，奥马哈]，
州：[伊利诺伊州，佛罗里达州，德克萨斯州，内布拉斯加州]}）
 
 
 
 df = pd.concat（[df1，df2]）
 df = df.reset_index（drop = True）
 
 df_gpby = df.groupby list（df.columns））
 idx = [x [0] for x in df_gpby.groups.values（）if len（x）== 1] 
 blah = df.reindex（idx）
  

解决方案

基于@ EdChum的建议：

  df = pd.merge（df1，df2，how ='outer'，indicator = True）
 rows_in_df1_not_in_df2 = df [df ['_ merge '] =='left_只有'] [df1.columns] 
 
 rows_in_df1_not_in_df2 
 
 |索引|城市|州| 
 | ------ | ------------ | ------------ | 
 | 1 |圣弗朗西斯|加利福尼亚| 
 | 2 |波士顿|马萨诸塞| 
  

I have 2 dataFrames and want to compare them and return rows from the first one (df1) that are not in the second one (df2). I found a way to compare them and return the differences, but can't figure out how to return only missing ones from df1.

import pandas as pd
from pandas import Series, DataFrame

df1 = pd.DataFrame( { 
"City" : ["Chicago", "San Franciso", "Boston"] , 
"State" : ["Illinois", "California", "Massachusett"] } )

df2 = pd.DataFrame( { 
"City" : ["Chicago",  "Mmmmiami", "Dallas" , "Omaha"] , 
"State" : ["Illinois", "Florida", "Texas", "Nebraska"] } )



df = pd.concat([df1, df2])
df = df.reset_index(drop=True)

df_gpby = df.groupby(list(df.columns))
idx = [x[0] for x in df_gpby.groups.values() if len(x) == 1]
blah = df.reindex(idx)

解决方案

Building on @EdChum's suggestion:

df = pd.merge(df1, df2, how='outer', indicator=True)
rows_in_df1_not_in_df2 = df[df['_merge']=='left_only'][df1.columns]

rows_in_df1_not_in_df2

|Index |City        |State       |
|------|------------|------------|
|1     |San Franciso|California  |
|2     |Boston      |Massachusett|

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