Python Pandas df.ix 未按预期执行

Question

这是我的功能:

 def clean_zipcodes(df):
    df.ix[df['workCountryCode'].str.contains('USA') & \
    df['workZipcode'].astype(str).str.len() > 5, 'workZipcode'] = \
    df['workZipcode'].astype(int).floordiv(10000)


df.ix[df['contractCountryCode'].str.contains('USA') & \
    df['contractZipcode'].astype(str).str.len() > 5, 'contractZipcode'] = \
    df['contractZipcode'].astype(int).floordiv(10000)

return df

这是我期望的测试函数:

def test_clean_zipcodes():
testDf = pandas.DataFrame({'unique_transaction_id'  : ['1', '1', '1'],
                           'workZipcode'            : [838431000, 991631000, 99163],
                           'contractZipcode'        : [838431000, 991631000, 99163],
                           'workCountryCode'        : ['USA: STUFF', 'NONE: STUFF', 'USA: STUFF'],
                           'contractCountryCode'    : ['USA: STUFF', 'NONE: STUFF', 'USA: STUFF']})

resultDf = pandas.DataFrame({'unique_transaction_id'    : ['1', '1', '1'],
                              'workZipcode'             : [83843, 991631000, 99163],
                              'contractZipcode'         : [83843, 991631000, 99163],
                              'workCountryCode'         : ['USA: STUFF', 'NONE: STUFF', 'USA: STUFF'],
                              'contractCountryCode'     : ['USA: STUFF', 'NONE: STUFF', 'USA: STUFF']})

assert resultDf.equals(clean_zipcodes(testDf))

除了缩进不正确(没有转换为 SO 格式)之外，df.ix 没有按预期执行。它不会对 contractZipcode 或 workZipcode 列执行任何转换。第一行应更改为 83843，如 resultDf 中所述。

提前致谢!

Answer 1

In [2]: import pandas as pd


In [3]: testDf = pd.DataFrame({'unique_transaction_id'  : ['1', '1', '1'],
   ...:                            'workZipcode'            : [838431000, 991631000, 99163],
   ...:                            'contractZipcode'        : [838431000, 991631000, 99163],
   ...:                            'workCountryCode'        : ['USA: STUFF', 'NONE: STUFF', 'USA: STUFF'],
   ...:                            'contractCountryCode'    : ['USA: STUFF', 'NONE: STUFF', 'USA: STUFF']}
   ...:                      )
   ...:
   ...: resultDf = pd.DataFrame({'unique_transaction_id'    : ['1', '1', '1'],
   ...:                               'workZipcode'             : [83843, 991631000, 99163],
   ...:                               'contractZipcode'         : [83843, 991631000, 99163],
   ...:                               'workCountryCode'         : ['USA: STUFF', 'NONE: STUFF', 'USA: STUFF'],
   ...:                               'contractCountryCode'     : ['USA: STUFF', 'NONE: STUFF', 'USA: STUFF']})
   ...:
   ...:
   ...:

请注意，当您尝试像这样建立索引时，会返回一个空切片:

In [4]: testDf.ix[testDf['workCountryCode'].str.contains('USA') &
                  testDf['workZipcode'].astype(str).str.len() > 5,
                  'workZipcode']
Out[4]: Series([], Name: workZipcode, dtype: int64)

如果您在不同的过滤器周围添加括号:

In [5]: testDf.ix[(testDf['workCountryCode'].str.contains('USA'))
                  & (testDf['workZipcode'].astype(str).str.len() > 5),
                  'workZipcode']
Out[5]:
0    838431000
Name: workZipcode, dtype: int64

你会得到你想要的。如果您使用 loc 也没关系:

In [6]: testDf.loc[testDf['workCountryCode'].str.contains('USA') &
                   testDf['workZipcode'].astype(str).str.len() > 5, 
                  'workZipcode']
Out[6]: Series([], Name: workZipcode, dtype: int64)

所以这是清理过的函数: 为了便于阅读，我添加了一些小的 lambda。

In [7]: def clean_zipcodes_loc(df):
   ...:     strlen = lambda x: x.astype(str).str.len()
   ...:     floordiv = lambda x: x.astype(int).floordiv(10000)
   ...:
   ...:     df.loc[((strlen(df.workZipcode)) > 5) &
   ...:            df.workCountryCode.str.contains("USA"),
   ...:           'workZipcode'] = floordiv(df.workZipcode)
   ...:
   ...:     df.loc[((strlen(df.contractZipcode)) > 5) &
   ...:            df.contractCountryCode.str.contains("USA"),
   ...:           'contractZipcode'] = floordiv(df.contractZipcode)
   ...:
   ...:     return df
   ...:

In [8]: clean_zipcodes_loc(testDf) == resultDf
Out[8]:
  contractCountryCode contractZipcode unique_transaction_id workCountryCode  \
0                True            True                  True            True
1                True            True                  True            True
2                True            True                  True            True

  workZipcode
0        True
1        True
2        True