Question

我需要比较数据帧中的两个连续行，例如：

df：

time        door       name
00:01:10    in         alex
00:01:10    in         alex
02:01:10    out        alex
03:01:10    in         alex
04:01:10    out        alex
04:01:10    out        alex

如果door是连续两行中的in（或out），则需要删除重复项。

这是我的代码的一部分：

import pandas as pd

file_name='test.xlsx'

df = pd.read_excel(file_name, header=0, index= False)
mydf = df.sort_values(by='time')
for i in range (len(mydf)):
    if (mydf[['door']] != mydf[['door']].shift(-1)).any(axis=1):
        print('ok')
    else:
        print ('nok')

我收到此错误：

if ((mydf[['Door Name']] != mydf[['Door Name']].shift(-1).any(axis=1))):

  File "C:\Users\khou\AppData\Local\Continuum\anaconda3\lib\site-packages\pandas\core\generic.py", line 1478, in __nonzero__
    .format(self.__class__.__name__))

ValueError: The truth value of a DataFrame is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

我不知道如何解决它，任何帮助将不胜感激。

Answer 1

您可以先使用np.where，然后再使用dropna

df['door'] = np.where((df['door'] == df['door'].shift(-1)), np.nan, df['door'])

df.dropna(how='any', axis=0, inplace= True)

print(df)

     time door  name
 00:01:10   in  alex
 02:01:10  out  alex
 03:01:10   in  alex
 04:01:10  out  alex

或

并且如果door.values总是有重复的时间，那么您可以简单地将df.drop_duplicates与参数keep = 'first' and subset = ['time', 'door']一起使用

df.drop_duplicates(subset=['time', 'door'], keep='first', inplace= True)

print(df)

     time door  name
 00:01:10   in  alex
 02:01:10  out  alex
 03:01:10   in  alex
 04:01:10  out  alex

Answer 2

您可以使用以下方法首先获取索引，然后根据索引删除：

代码

import pandas as pd

mydf=pd.DataFrame({'time':['00:01:10','00:01:10','02:01:10','03:01:10','04:01:10','04:01:10'],
                'door':['in','in','out','in','out','out'],
               'name':['alex','alex','alex','alex','alex','alex']})

idx=[]
for i in range (0,len(mydf)):
    if i == 0:
        print ('index '+str(i)+' ok')
    elif mydf['door'][i] != mydf['door'][i-1]:
        print('index '+str(i)+' ok')
    else:
        print ('index '+str(i)+' nok')
        idx.append(i)

mydf.drop(mydf.index[[idx]],inplace=True)

print('\n',mydf)

输出

index 0 ok
index 1 nok
index 2 ok
index 3 ok
index 4 ok
index 5 nok

        time door  name
0  00:01:10   in  alex
2  02:01:10  out  alex
3  03:01:10   in  alex
4  04:01:10  out  alex

比较和删除数据框Python中的连续行

2 个答案: