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python之将行与引用行 pandas 进行比较

2025年05月04日23softidea

我有一个示例数据集:

import pandas as pd 
 
df = { 
'rank1':[1,2,3,4,5,6,7,8], 
'rank12':[1,2,3,4,8,9,37,15], 
'rank13':[1,2,3,4,12,6,24,14], 
'N':['','','','','','','',''], 
'code#':[1945, 13060, 610, 402, 1067, 180, 411, 93], 
'score1':[100,97,95,92,87,85,80,79], 
'score2':['yes','yes','no','no','yes','yes','no','yes'], 
'score3':[10,9,10,9,9,8,9,9], 
'score4':['yes','yes','no','no','yes','yes','no','yes'], 
'score5':[2,3,2,2,2,2,2,2] 
} 
df = pd.DataFrame(df) 

看起来像:

df 
Out[130]:  
  N   code#   rank1   rank12  rank13  score1 score2  score3 score4  score5 
0     1945      1       1       1     100    yes      10    yes       2 
1    13060      2       2       2      97    yes       9    yes       3 
2      610      3       3       3      95     no      10     no       2 
3      402      4       4       4      92     no       9     no       2 
4     1067      5       8      12      87    yes       9    yes       2 
5      180      6       9       6      85    yes       8    yes       2 
6      411      7      37      24      80     no       9     no       2 
7       93      8      15      14      79    yes       9    yes       2 

我想将代码# = 93 的最后一行与此处的其余行(仅数字列)进行比较。如果任何值 < 最后一行,将该值替换为 1,如果 >= 最后一行,将该值替换为 0。

期望的输出:

    Out[130]:  
  N   code#   rank1   rank12  rank13  score1 score2  score3 score4  score5 
0     1945      1       1       1      0     yes      0     yes       0 
1    13060      1       1       1      0     yes      0     yes       0 
2      610      1       1       1      0     no       0     no        0 
3      402      1       1       1      0     no       0     no        0 
4     1067      1       1       1      0     yes      0     yes       0 
5      180      1       1       1      0     yes      1     yes       0 
6      411      1       0       0      0     no       0     no        0 
7       93      8      15      14      79    yes       9    yes       0 

我的想法: 1.创建一个字典,以列名作为键,最后一行的值作为值 2.遍历每一行并与字典值进行比较

我的尝试:

baserow = df[df['code#'] == 93]  #get the last row 
dict=baserow.to_dict(orient='list') #make the last row into a dictionary  
 
try:  #i'm using a try except here because there are non-numeric columns here, this will not raise errors. 
for index, row in df.iterrows(): #iterating through each row 
    for key, value in dict.items():  #iterating through the dictionary 
        Othervals=df.ix[index][key]  #individual value for compare data 
        vals = dict.get(key) 
        vals= vals[0]  #get dictionary value 
        if vals>Othervals:  #if the dictionary value > other row value then make the cell 1, else 0 
            df[[index][key]] == 1    
        else: 
            df[[index][key]] == 0 
except: 
    pass 

但 df 没有改变,它仍然具有相同的旧值。

请您参考如下方法:

借用@Psidom 的df_numeric

df_numeric = df.drop('code#',1).select_dtypes(include=[pd.np.number])    
 
v = df_numeric.values 
df.loc[df.index[:-1], df_numeric.columns] = np.where(v[:-1] < v[-1], 1, 0) 
 
df 
 
  N  code#  rank1  rank12  rank13  score1 score2  score3 score4  score5 
0     1945      1       1       1       0    yes       0    yes       0 
1    13060      1       1       1       0    yes       0    yes       0 
2      610      1       1       1       0     no       0     no       0 
3      402      1       1       1       0     no       0     no       0 
4     1067      1       1       1       0    yes       0    yes       0 
5      180      1       1       1       0    yes       1    yes       0 
6      411      1       0       0       0     no       0     no       0 
7       93      8      15      14      79    yes       9    yes       2