Python 当列表包含其他列表中的特定字符时，请删除列表值

您好，我有一个名为“d”的dict和一个“待删除”列表，其中列表中的值如下所示：

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']

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)

我需要从列表d中删除那些包含字符串“ZZ”和“PRE”的列表值。因此，我的输出列表应仅为值：

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

---

如何在python中做到这一点？

您可以使用如下列表理解：

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:
Out[14]:

['ZZ', 'PRE']

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)

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dict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()}

您可以使用函数和列表生成/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

或仅一行-

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        ']}

---

您可以使用

re.sub

，

列表理解

和

听写理解

：

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        ']}

---

您可以使用正则表达式。>您不需要在

任何

参数中使用

[]

。此外，您可以将其设置为dict理解，这样它也可以用于包含多个元素的词典。除此之外，这正是我想发布的内容。是的，我还在玩弄听写理解，但觉得它开始看起来太乱了！我想我现在再看一眼