Question

嗨，我有一个名为＆＃39; d＆＃39;和一个清单＆＃39; to_be_removed＆＃39;其中列表中的值如下：

d:

{'data': [' VL0002511CA       000000000682414100000000000110000        ',
 ' VL0002511ZZ       000000000682414100000000000110000        ',
 ' VL0002512  PRE1985000000000682414100000000000110000        ',
 ' VL0002521CA       000000001888990100000000000310000        ',
 ' VL0002521ZZ       000000001888990100000000000310000        ',
 ' VL0002522  PRE1985000000001888990100000000000310000        ',
 ' VL0002531CA       000000001223831100000000000210000        ',
 ' VL0002531ZZ       000000001223831100000000000210000        ',
 ' VL0002532  PRE1985000000001223831100000000000210000        ',
 ' VL0007871CA       000000001463787100000000000210000        ',
 ' VL0007871ZZ       000000001463787100000000000210000        ']}

to_be_removed:
Out[14]:

['ZZ', 'PRE']

我需要删除那些包含字符串＆＃39; ZZ＆＃39;和＆＃39; PRE＆＃39;从列表中d。所以我的输出列表应该只是值：

final_list:
{'data': [' VL0002511CA       000000000682414100000000000110000        ',
 ' VL0002521CA       000000001888990100000000000310000        ',
 ' VL0002531CA       000000001223831100000000000210000        ',
 ' VL0007871CA       000000001463787100000000000210000        ']}

我怎么能在python中做到这一点？

Answer 1

您可以按如下方式使用列表推导：

d = {'data': [' VL0002511CA       000000000682414100000000000110000        ',
 ' VL0002511ZZ       000000000682414100000000000110000        ',
 ' VL0002512  PRE1985000000000682414100000000000110000        ',
 ' VL0002521CA       000000001888990100000000000310000        ',
 ' VL0002521ZZ       000000001888990100000000000310000        ',
 ' VL0002522  PRE1985000000001888990100000000000310000        ',
 ' VL0002531CA       000000001223831100000000000210000        ',
 ' VL0002531ZZ       000000001223831100000000000210000        ',
 ' VL0002532  PRE1985000000001223831100000000000210000        ',
 ' VL0007871CA       000000001463787100000000000210000        ',
 ' VL0007871ZZ       000000001463787100000000000210000        ']}

to_be_removed = ['PRE', 'ZZ']
updated = {'data': [item for item in d['data'] if not any(subst in item for subst in to_be_removed)] }
print(updated)

<强>输出

{'data': [' VL0002511CA       000000000682414100000000000110000        ', 
' VL0002521CA       000000001888990100000000000310000        ', 
' VL0002531CA       000000001223831100000000000210000        ', 
' VL0007871CA       000000001463787100000000000210000        ']}

要允许使用多个键的词典，请使用词典理解：

updated = {k: [item for item in v if not any(subst in item for subst in to_be_removed)] for k, v in d.items()}

Answer 2

您可以使用all功能和列表生成/理解。

d={'data': [' VL0002511CA       000000000682414100000000000110000        ',
          ' VL0002511ZZ       000000000682414100000000000110000        ',
          ' VL0002512  PRE1985000000000682414100000000000110000        ',
          ' VL0002521CA       000000001888990100000000000310000        ',
          ' VL0002521ZZ       000000001888990100000000000310000        ',
          ' VL0002522  PRE1985000000001888990100000000000310000        ',
          ' VL0002531CA       000000001223831100000000000210000        ',
          ' VL0002531ZZ       000000001223831100000000000210000        ',
          ' VL0002532  PRE1985000000001223831100000000000210000        ',
          ' VL0007871CA       000000001463787100000000000210000        ',
          ' VL0007871ZZ       000000001463787100000000000210000        ']}

to_be_removed=['ZZ', 'PRE']
lst = []
for j in d['data']:
    if all(not i in j for i in to_be_removed):
        lst.append(j)

final_data = {'data': lst}
print final_data

或只是一个班次 -

print {'data': [j for j in d['data'] if all(not i in j for i in to_be_removed)]}

输出 -

{'data': [' VL0002511CA       000000000682414100000000000110000        ', ' VL0002521CA       000000001888990100000000000310000        ', ' VL0002531CA       000000001223831100000000000210000        ', ' VL0007871CA       000000001463787100000000000210000        ']}

Answer 3

您可以使用re.sub，list comprehension和dict comprehension执行此操作：

In [182]: {'data' : [l2 for l2 in [re.sub('.*[(ZZ)|(PRE)]+.*', '', l1) for l1 in d['data']] if l2]}
Out[182]:
{'data': [' VL0002511CA       000000000682414100000000000110000        ',
  ' VL0002521CA       000000001888990100000000000310000        ',
  ' VL0002531CA       000000001223831100000000000210000        ',
  ' VL0007871CA       000000001463787100000000000210000        ']}

当它包含来自其他列表python的特定字符时，删除列表值

3 个答案: