TypeError:'Series'对象是可变的,因此它们不能被散列

问题描述 投票:0回答:2

我知道这个错误是常见的,我尝试了一些解决方案,我抬起头来仍然无法理解有什么问题。我想这是由于row和row1的可变形式,但我无法弄明白

我想做什么?我有2个数据帧。我需要迭代第一个1的行,并且对于第一个的每一行迭代第二个并检查某些列的单元格值。我的代码和不同的尝试:

a=0
b=0
  for row in Correction.iterrows():
        b+=1
        for row1 in dataframe.iterrows():
            c+=1
            a=0
            print('Handling correction '+str(b)+' and deal '+str(c))
            if (Correction.loc[row,['BO Branch Code']]==dataframe.loc[row1,['wings Branch']] and Correction.loc[row,['Profit Center']]==dataframe.loc[row1,['Profit Center']] and Correction.loc[row,['Back Office']]==dataframe.loc[row1,['Back Office']]
                and Correction.loc[row,['BO System Code']]==dataframe.loc[row1,['BO System Code']]):

我也试过了

a=0
b=0
 for row in Correction.iterrows():
        b+=1
        for row1 in dataframe.iterrows():
            c+=1
            a=0
            print('Handling correction '+str(b)+' and deal '+str(c))
            if (Correction[row]['BO Branch Code']==dataframe[row1]['wings Branch'] and Correction[row]['Profit Center']==dataframe[row1]['Profit Center'] and Correction[row]['Back Office']==dataframe[row1]['Back Office']
                and Correction[row]['BO System Code']==dataframe[row1]['BO System Code']):

a=0
b=0
 for row in Correction.iterrows():
        b+=1
        for row1 in dataframe.iterrows():
            c+=1
            a=0
            print('Handling correction '+str(b)+' and deal '+str(c))
            if (Correction.loc[row,['BO Branch Code']]==dataframe[row1,['wings Branch']] and Correction[row,['Profit Center']]==dataframe[row1,['Profit Center']] and Correction[row,['Back Office']]==dataframe[row1,['Back Office']]
                and Correction[row,['BO System Code']]==dataframe[row1,['BO System Code']]):
python pandas hash immutability
2个回答
0
投票

我通过改变我的for循环找到了解决办法,现在我的代码是:

a=0
b=0
 for index in Correction.index:
        b+=1
        for index1 in dataframe.index:
            c+=1
            a=0
            print('Handling correction '+str(b)+' and deal '+str(c))
            if (Correction.loc[row,'BO Branch Code']==dataframe.loc[row1,'Wings Branch]] and Correction.loc[row,'Profit Center']==dataframe.loc[row1,'Profit Center'] and Correction.loc[row,'Back Office']==dataframe.loc[row1,'Back Office']
                and Correction.loc[row,'BO System Code']==dataframe.loc[row1,'BO System Code']):

0
投票

我认为你正在迭代你的df错误

for row in Correction.itertuples():
    bo_branch_code = row['BO Branch Code']
    for row1 in dataframe.itertuples():
        if row1['wings Branch'] == bo_branch_code:
            # do stuff here

参考如何迭代DataFrame:https://github.com/vi3k6i5/pandas_basics/blob/master/2.A%20Iterate%20over%20a%20dataframe.ipynb

我计算了你的索引方法和迭代方法。结果如下:

import pandas as pd
import numpy as np
import time

df = pd.DataFrame(np.random.randint(0,100,size=(10, 4)), columns=list('ABCD'))

df_2 = pd.DataFrame(np.random.randint(0,100,size=(10, 4)), columns=list('ABCD'))

def test_time():
    for index in df.index:
        for index1 in df_2.index:
            if (df.loc[index, 'A'] == df_2.loc[index1, 'A']):
                continue

def test_time_2():
    for idx, row in df.iterrows():
        a_val = row['A']
        for idy, row_1 in df_2.iterrows():
            if (a_val == row_1['A']):
                continue

start= time.clock()
test_time()
end= time.clock()
print(end-start)
# 0.038514999999999855

start= time.clock()
test_time_2()
end= time.clock()
print(end-start)
# 0.009272000000000169

简单地说iterrows比你的方法更快。

参考循环数据框What is the most efficient way to loop through dataframes with pandas?的好方法

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