我有一个excel文件，我将其作为数据帧导入。我想循环遍历数据帧的列。例如，我想将第二列与第一列进行比较，然后将第三列与第二列进行比较。我已将rule_id列转换为索引。这是数据：
rule_id reqid1 reqid2 reqid3
53139 0 0 1
51181 1 1 0
50412 0 1 1
50356 0 0 1
50239 0 1 0
50238 1 1 0
50014 1 0 1
这就是我正在使用的代码。
for n in fin2.columns[0:]:
n = 0
n_int = int(n)
if ([fin2.iloc[: , n_int+1] != fin2.iloc[: , n_int]]):
print dframe2
if ([fin2.iloc[: , n_int+1] == fin2.iloc[: , n_int]]):
print dframe3
n = n+1
有了这段代码，我只能将第二列与第一列进行比较，我将n的值设置为0，并应用了逻辑，n=n+1，每当条件满足时，该逻辑就会增加n的值。非常感谢您的帮助。我创建了以下两个功能：
def solved_prior(df):
n = 0
n_int = int(n)
df['solved_prior'] = np.where(df.iloc[: , n_int+1] < df.iloc[: , n_int] , 100 , np.nan)
return df
和
def repeated_prior(df):
n = 0
n_int = int(n)
df['repeated_prior'] = np.where((df.iloc[: , n_int+1] == df.iloc[: , n_int]) & (df.iloc[: , n_int] == 1) , 1 , np.nan)
return df
我已经将这些函数分别存储在daframe2和dataframe3中。我想要第二列和第一列之间的第一次比较结果为：
rule_id reqid1 reqid2 reqid3 solved prior repeated prior
53139 0 0 1 NaN NaN
51181 1 1 0 NaN 1
50412 0 1 1 NaN NaN
50356 0 0 1 NaN NaN
50239 0 1 0 NaN NaN
50238 1 1 0 NaN 1
50014 1 0 1 100 NaN
第3列和第2列之间的比较结果应该是这样的：
rule_id reqid1 reqid2 reqid3 solved prior repeated prior
53139 0 0 1 NaN NaN
51181 1 1 0 100 NaN
50412 0 1 1 NaN 1
50356 0 0 1 NaN NaN
50239 0 1 0 100 NaN
50238 1 1 0 NaN NaN
50014 1 0 1 NaN NaN
就像注释状态一样，您的预期输出可能会影响最佳解决方案。请记住，在列上循环很少是最佳的解决方案。我建议简单地添加新的列，指示被比较的列是否相等。例如：
In [1]: import pandas as pd
In [2]: df = pd.DataFrame({'rule_id': [53139,51181,50412,50356,50239,50238,50014], 'reqid1':[0,1,0,0,0,1,1],'reqid2':[0,1,1,0,1,1,0],'reqid3':[1,0,1,1,0,0,1]})
In [3]: df
Out[3]:
rule_id reqid1 reqid2 reqid3
0 53139 0 0 1
1 51181 1 1 0
2 50412 0 1 1
3 50356 0 0 1
4 50239 0 1 0
5 50238 1 1 0
6 50014 1 0 1
In [4]: df['compare_1_and_2'] = df.reqid1 == df.reqid2
In [5]: df
Out[5]:
rule_id reqid1 reqid2 reqid3 compare_1_and_2
0 53139 0 0 1 True
1 51181 1 1 0 True
2 50412 0 1 1 False
3 50356 0 0 1 True
4 50239 0 1 0 False
5 50238 1 1 0 True
6 50014 1 0 1 False
In [6]: df['compare_2_and_3'] = df.reqid2 == df.reqid3
In [7]: df
Out[7]:
rule_id reqid1 reqid2 reqid3 compare_1_and_2 compare_2_and_3
0 53139 0 0 1 True False
1 51181 1 1 0 True False
2 50412 0 1 1 False True
3 50356 0 0 1 True False
4 50239 0 1 0 False False
5 50238 1 1 0 True False
6 50014 1 0 1 False False
现在，如果列很长，您可能会发现any()和all()很有用。查看是否有任何值为真(至少有一个值相同)：
In [8]: df.compare_1_and_2.any()
Out[8]: True
查看所有值是否为真(列相同)：
In [9]: df.compare_1_and_2.all()
Out[9]: False
编辑：(以匹配预期输出)现在，使用布尔列来匹配您需要的就很简单了
df['solved_prior_1_vs_2'] = np.NaN
df['repeated_prior_1_vs_2'] = np.NaN
df.loc[(df.compare_1_and_2 == False) & (df.reqid1 == 1),'solved_prior_1_vs_2'] = 100
df.loc[(df.compare_1_and_2 == True) & (df.reqid1 == 1),'repeated_prior_1_vs_2'] = 1
结果如下：
In [27]: df[['rule_id','reqid1','reqid2','solved_prior_1_vs_2','repeated_prior_1_vs_2']]
Out[27]:
rule_id reqid1 reqid2 solved_prior_1_vs_2 repeated_prior_1_vs_2
0 53139 0 0 NaN NaN
1 51181 1 1 NaN 1.0
2 50412 0 1 NaN NaN
3 50356 0 0 NaN NaN
4 50239 0 1 NaN NaN
5 50238 1 1 NaN 1.0
6 50014 1 0 100.0 NaN
现在，您可以删除不需要的列，并对2和3进行比较。也可以将新列转换为int。
最终编辑(希望如此)：一个更简单的解决方案是只定义一个函数，例如：
def compare_columns(df, col1, col2):
repeated_name = "{}_{}_repeated".format(col1, col2)
solved_name = "{}_{}_solved".format(col1, col2)
diff = df[col1] == df[col2]
col1_is_1 = df[col1] == 1
df[repeated_name] = 100
df[solved_name] = 1
df[repeated_name] = df[repeated_name].astype(int)
df[solved_name] = df[sovled_name].astype(int)
df.loc[~(diff & col1_is_1), solved_name] = np.NaN
df.loc[~(~diff & col1_is_1), repeated_name] = np.NaN
return df
现在你可以做：
In [42]: df1 = compare_columns(df, 'reqid1', 'reqid2')
In [43]: df1
Out[43]:
rule_id reqid1 reqid2 reqid3 reqid1_reqid2_repeated reqid1_reqid2_solved
0 53139 0 0 1 NaN NaN
1 51181 1 1 0 NaN 1
2 50412 0 1 1 NaN NaN
3 50356 0 0 1 NaN NaN
4 50239 0 1 0 NaN NaN
5 50238 1 1 0 NaN 1
6 50014 1 0 1 100 NaN