当特定类型的值出现在列中时，熊猫删除一行

Question

我有这样的 DF

         UNIT  EXITSn_hourly           Interval
1867     R081            104  00:00:00-04:00:00
1868     R081              0  04:00:00-04:00:00
1869     R081            129  04:00:00-08:00:00
1870     R081            521  08:00:00-12:00:00
1871     R081           1048  12:00:00-16:00:00
2838     R032             38  00:00:00-04:00:00
2839     R032              0  04:00:00-04:00:00
2840     R032             89  04:00:00-08:00:00
2841     R032            470  08:00:00-12:00:00

当 Interval 具有这种特定格式时，我需要删除整行

1868     R081              0  04:00:00-04:00:00

我不仅想删除04:00:00-04:00:00，还想删除类似的值，例如

01:00:00-01:00:00

其实这是我原来的df。我创建了一个间隔

    C/A  UNIT       SCP     DATEn     TIMEn    DESCn  ENTRIESn   EXITSn
0  A002  R051  02-00-00  06-29-13  00:00:00  REGULAR   4174592  1433672
1  A002  R051  02-00-00  06-29-13  04:00:00  REGULAR   4174628  1433675
2  A002  R051  02-00-00  06-29-13  08:00:00  REGULAR   4174641  1433706
3  A002  R051  02-00-00  06-29-13  12:00:00  REGULAR   4174741  1433775
4  A002  R051  02-00-00  06-29-13  16:00:00  REGULAR   4174936  1433826
5  A002  R051  02-00-00  06-29-13  20:00:00  REGULAR   4175270  1433877
6  A002  R051  02-00-00  06-30-13  00:00:00  REGULAR   4175403  1433908
7  A002  R051  02-00-00  06-30-13  04:00:00  REGULAR   4175441  1433914
8  A002  R051  02-00-00  06-30-13  08:00:00  REGULAR   4175457  1433928
9  A002  R051  02-00-00  06-30-13  12:00:00  REGULAR   4175520  1433981

我使用此代码创建了间隔

import copy

df = copy.deepcopy(turnstile_data)
pdf = df.shift(periods=1)

df['ENTRIESn_hourly'] = df['ENTRIESn'] - pdf['ENTRIESn'].fillna(0)
df['EXITSn_hourly'] = df['EXITSn'] - pdf['EXITSn'].fillna(0)
df['Interval'] = pdf['TIMEn']+'-'+ df['TIMEn'].fillna(0)
df.loc[(df['ENTRIESn'] == 0), 'ENTRIESn_hourly'] = 0
df.loc[(df['EXITSn'] == 0), 'EXITSn_hourly'] = 0
df.loc[(df['C/A'] != pdf['C/A']) | (df['UNIT'] != pdf['UNIT']) | (df['SCP'] != pdf['SCP']), ['ENTRIESn_hourly', 'EXITSn_hourly','Interval']] = 0

df = df[df.Interval != 0]
print df.head(20)

head7=copy.deepcopy(df)
required_df=head7[['UNIT','EXITSn_hourly','Interval']].groupby(head7.UNIT)
print required_df.head(5)

Answer 1

您可以比较部分字符串，然后按子集删除它们：

print df.Interval.str[0:2]
1867    00
1868    04
1869    04
1870    08
1871    12
2838    00
2839    04
2840    04
2841    08
Name: Interval, dtype: object

print df.Interval.str[0:2] != df.Interval.str[9:11]
1867     True
1868    False
1869     True
1870     True
1871     True
2838     True
2839    False
2840     True
2841     True
Name: Interval, dtype: bool

print df[df.Interval.str[0:2] != df.Interval.str[9:11]]
      UNIT  EXITSn_hourly           Interval
1867  R081            104  00:00:00-04:00:00
1869  R081            129  04:00:00-08:00:00
1870  R081            521  08:00:00-12:00:00
1871  R081           1048  12:00:00-16:00:00
2838  R032             38  00:00:00-04:00:00
2840  R032             89  04:00:00-08:00:00
2841  R032            470  08:00:00-12:00:00

编辑：

我检查了您的代码，也许您可​​以省略 copy.deepcopy 并使用 copy：

df = turnstile_data.copy(deep=True)

df['ENTRIESn_hourly'] = (df['ENTRIESn'] - df['ENTRIESn'].shift(periods=1)).fillna(0)
df['EXITSn_hourly'] = (df['EXITSn'] - df['EXITSn'].shift(periods=1)).fillna(0)
df['Interval'] = (df['TIMEn'].shift(periods=1)+'-'+ df['TIMEn']).fillna(0)

df.loc[(df['ENTRIESn'] == 0), 'ENTRIESn_hourly'] = 0
df.loc[(df['EXITSn'] == 0), 'EXITSn_hourly'] = 0
df.loc[(df['C/A'] != df['C/A'].shift(periods=1)) | 
       (df['UNIT'] != df['UNIT'].shift(periods=1)) | 
       (df['SCP'] != df['SCP'].shift(periods=1)), 
['ENTRIESn_hourly', 'EXITSn_hourly','Interval']] = 0

print df.head(5)
   ENTRIESn_hourly  EXITSn_hourly           Interval  
0                0              0                  0  
1               36              3  00:00:00-04:00:00  
2               13             31  04:00:00-08:00:00  
3              100             69  08:00:00-12:00:00  
4              195             51  12:00:00-16:00:00  

required_df=df[['UNIT','EXITSn_hourly','Interval']].groupby(df.UNIT)

print required_df.head(5)
   UNIT  EXITSn_hourly           Interval
0  R051              0                  0
1  R051              3  00:00:00-04:00:00
2  R051             31  04:00:00-08:00:00
3  R051             69  08:00:00-12:00:00
4  R051             51  12:00:00-16:00:00

Answer 2

可能您想将 Interval 拆分为 Interval_start 和 Interval_end 并检查它们是否相等：

df['Interval_start'] = df['Interval'].map(lambda s: s.split('-')[0])
df['Interval_end'] = df['Interval'].map(lambda s: s.split('-')[1])
df.query("Interval_start != Interval_end")

      UNIT  EXITSn_hourly           Interval Interval_start Interval_end
1867  R081            104  00:00:00-04:00:00       00:00:00     04:00:00
1869  R081            129  04:00:00-08:00:00       04:00:00     08:00:00
1870  R081            521  08:00:00-12:00:00       08:00:00     12:00:00
1871  R081           1048  12:00:00-16:00:00       12:00:00     16:00:00
2838  R032             38  00:00:00-04:00:00       00:00:00     04:00:00
2840  R032             89  04:00:00-08:00:00       04:00:00     08:00:00
2841  R032            470  08:00:00-12:00:00       08:00:00     12:00:00