python中带重复项的减法列表

Question

我有两张单子：

l1=['TUDCAPL.CLM_NUM_CD',
    'TUDCAPL.CRT_TS',
    'TUDCAPL.CLM_NUM_CD',
    'TUDCAPL.CRT_TS',
    'TUDCAPJ.CLM_NUM_CD',
    'TUDCAPJ.CRT_TS']

l2 = ['TUDCAPL.CLM_NUM_CD',
      'TUDCAPL.CRT_TS']

我希望我的结果是

l3 = ['TUDCAPL.CLM_NUM_CD',
      'TUDCAPL.CRT_TS',
      'TUDCAPJ.CLM_NUM_CD',
      'TUDCAPJ.CRT_TS']

我用了l3 = [x for x in l1 if x not in l2] 但是结果是['TUDCAPJ.CLM_NUM_CD','TUDCAPJ.CRT_TS']忽略了重复项。我怎样才能获得与其他唯一值一起的副本？你知道吗

Answer 1

复制列表（如果您不想更改l1）。在l2中的项上循环一次，并将其从新列表中删除。remove函数将丢弃每个项的第一次出现，而不影响其他项和顺序。你知道吗

l1 = ['TUDCAPL.CLM_NUM_CD',
      'TUDCAPL.CRT_TS',
      'TUDCAPL.CLM_NUM_CD',
      'TUDCAPL.CRT_TS',
      'TUDCAPJ.CLM_NUM_CD',
      'TUDCAPJ.CRT_TS']

l2 = ['TUDCAPL.CLM_NUM_CD',
      'TUDCAPL.CRT_TS']

l3 = list(l1)
for x in l2:
    if x in l3:
        l3.remove(x)

from pprint import pprint
pprint(l3)

结果：

['TUDCAPL.CLM_NUM_CD', 
 'TUDCAPL.CRT_TS', 
 'TUDCAPJ.CLM_NUM_CD', 
 'TUDCAPJ.CRT_TS']

Answer 2

您可以使用set：

l3 = set(l1)
l3.update(l2)
print l3

Answer 3

我知道这不是最好的解决方案，但它将保持您列表中的顺序和独特元素：

from collections import OrderedDict

l1=['TUDCAPL.CLM_NUM_CD',
    'TUDCAPL.CRT_TS',
    'TUDCAPL.CLM_NUM_CD',
    'TUDCAPL.CRT_TS',
    'TUDCAPJ.CLM_NUM_CD',
    'TUDCAPJ.CRT_TS']

l2 = ['TUDCAPL.CLM_NUM_CD',
      'TUDCAPL.CRT_TS']


d = OrderedDict(zip(l1,[None]*len(l1)))
d2 = OrderedDict(zip(l2,[None]*len(l2)))

d.update(d2)
print(d.keys()) # this will contain the resulting list of strings