如何根據另一個 DataFrame 中的列更新 Pandas DataFrame 中的列

Question

假設我有 2 個數據幀

df1 = pd.DataFrame({'name': ['Jack', 'Lucy', 'Mark'], 'age': [1, 2, 3]})
df2 = pd.DataFrame({'name': ['Jack', 'Mark'], 'age': [10, 11], 'address': ['addr1', 'addr2']})

我應該用什么操作讓df1變成

name    age    address
--------------------
Jack    10     addr1
Lucy    2      NaN
Mark    11     addr2

Answer 1

您可以合並兩個 df 然后替換缺失值：

df_out = df1.merge(df2,on=['name'],how='left')
df_out['age'] =  df_out.apply(lambda x : x['age_y'] if x['age_y']>0 else x['age_x'],axis = 1)
df_out[['name','age','address']]

輸出

| name   |   age | address   |
|:-------|------:|:----------|
| Jack   |    10 | addr1     |
| Lucy   |     2 | nan       |
| Mark   |    11 | addr2     |

Answer 2

使用DataFrame.combine_first按name列轉換為兩個DataFrame的索引：

df1 = df1.set_index('name') 
df2 = df2.set_index('name')

df1 = df2.combine_first(df1).reset_index()
print (df1)
   name address   age
0  Jack   addr1  10.0
1  Lucy     NaN   2.0
2  Mark   addr2  11.0

應更改第一個原始解決方案：

df1 = df1.set_index('name')
df2 = df2.set_index('name')
df1 = df1.reindex(df1.columns.union(df2.columns, sort=False), axis=1)

df1.update(df2)
df1 = df1.reset_index()
print (df1)
   name   age address
0  Jack  10.0   addr1
1  Lucy   2.0     NaN
2  Mark  11.0   addr2

或者在DataFrame.merge和DataFrame.combine_first使用左連接的解決方案：

#left join df2, if existing columns name is added _ to end
df = df1.merge(df2, on='name', how='left', suffixes=('','_'))

#filter columns names
new_cols = df.columns[df.columns.str.endswith('_')]

#remove last char from column names
orig_cols = new_cols.str[:-1]
#dictionary for rename
d = dict(zip(new_cols, orig_cols))

#filter columns and replace NaNs by new appended columns
df[orig_cols] = df[new_cols].rename(columns=d).combine_first(df[orig_cols])
#remove appended columns 
df = df.drop(new_cols, axis=1)
print (df)
   name   age address
0  Jack  10.0   addr1
1  Lucy   2.0     NaN
2  Mark  11.0   addr2

Answer 3

您可以使用 concat、drop_duplicates、sort_index 和 reset_index

df = pd.concat([df1,df2],ignore_index=False, sort=False).drop_duplicates(["name"], keep="last").sort_index().reset_index(drop=True)