python - 将行与引用行 pandas 进行比较

Question

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。

Answer 1

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