从df1和df2创建DataFrame，如果在df1的列”value”中不存在，则从df2中获取空值。

huangapple
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英文:

Create DataFrame from df1 and df2 and take empty value from df2 for column value if not exist in df1 column value

问题

需要从df2中提取如果在df1中缺失的frn，并创建一个新的DataFrame。

英文: 
  1. df1 = pd.DataFrame({'call_sign': ['ASD','BSD','CDSF','GFDFD','FHHH'],'frn':['123','124','','656','']})
  3. df2 = pd.DataFrame({'call_sign': ['ASD','CDSF','BSD','GFDFD','FHHH'],'frn':['1234','','124','','765']})

need to get a new df like

  1. df2 = pd.DataFrame({'call_sign': ['ASD','BSD','CDSF','GFDFD','FHHH'],'frn':['123','','124','656','765']})

I need to take frn from df2 if it's missing in df1 and create a new df

答案1

得分: 1

将空字符串替换为缺失值，并使用 DataFrame.set_index 与 DataFrame.fillna，因为需要按照 df2.call_sign 的顺序添加 DataFrame.reindex:

  1. df = (df1.set_index('call_sign').replace('', np.nan)
  2.           .fillna(df2.set_index('call_sign').replace('', np.nan))
  3.           .reindex(df2['call_sign']).reset_index())
  4. print(df)
  5.   call_sign  frn
  6. 0       ASD  123
  7. 1      CDSF  NaN
  8. 2       BSD  124
  9. 3     GFDFD  656
  10. 4      FHHH  765
英文:

Replace empty strings to missing values and use DataFrame.set_index with DataFrame.fillna, because need ordering like df2.call_sign add DataFrame.reindex:

  1. df = (df1.set_index('call_sign').replace('', np.nan)
  2.           .fillna(df2.set_index('call_sign').replace('', np.nan))
  3.          .reindex(df2['call_sign']).reset_index())
  4. print(df)
  5.   call_sign  frn
  6. 0       ASD  123
  7. 1      CDSF  NaN
  8. 2       BSD  124
  9. 3     GFDFD  656
  10. 4      FHHH  765

答案2

得分: 0

如果您要更新df2，可以使用布尔索引：

  1. # 是否为空字符串？
  2. m = df2['frn'].eq('')
  3. # 从df1的值更新这些行
  4. df2.loc[m, 'frn'] = df2.loc[m, 'call_sign'].map(df1.set_index('call_sign')['frn'])

更新后的df2：

  1.   call_sign   frn
  2. 0       ASD  1234
  3. 1      CDSF      
  4. 2       BSD   124
  5. 3     GFDFD   656
  6. 4      FHHH   765
英文:

If you want to update df2 you can use boolean indexing:

  1. # is frn empty string?
  2. m = df2['frn'].eq('')
  3. # update those rows from the value in df1
  4. df2.loc[m, 'frn'] = df2.loc[m, 'call_sign'].map(df1.set_index('call_sign')['frn'])

Updated df2:

  1.   call_sign   frn
  2. 0       ASD  1234
  3. 1      CDSF      
  4. 2       BSD   124
  5. 3     GFDFD   656
  6. 4      FHHH   765

答案3

得分: 0

  1. temp = df1.merge(df2, how='left', on='call_sign')
  2. df1['frn'] = temp['frn_x'].where(temp['frn_x'] != '', temp['frn_y'])
  1.     call_sign    frn
  2. 0  ASD         123
  3. 1  BSD         
  4. 2  CDSF        124
  5. 3  GFDFD       656
  6. 4  FHHH        765
英文: 
  1. temp = df1.merge(df2,how='left',on='call_sign')
  2. df1['frn']=temp.frn_x.where(temp.frn_x!='',temp.frn_y)
  1.     call_sign 	frn
  2. 0 	ASD 	    123
  3. 1 	BSD 	
  4. 2 	CDSF 	    124
  5. 3 	GFDFD 	    656
  6. 4 	FHHH 	    765

huangapple
  • 本文由 发表于 2023年2月16日 14:18:43
  • 转载请务必保留本文链接：https://go.coder-hub.com/75468512.html
  • dataframe
  • pandas
  • python
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