如何使用python比较列表中的元素并检查第一个列表元素是否包含在另一个列表的元素中

Question

我有两个列表l1和l2包含如下元素

l1 = ['a','b','c','d']
l2 = ['a is greater', 'f is greater', 'c is greater']

我想比较l1的元素，看看l2是否包含它们的元素。期望输出将是

f is greater

按照我尝试过的代码，

for i in l2:
for j in l2:
if j not in l1:
print(i)

但是我观察到的输出是

a is greater
a is greater
f is greater
f is greater
c is greater
c is greater

请帮助我知道我需要添加什么才能得到适当的输出。谢谢。

Answer 1

使用说明:

l1 = ['a','b','c','d']
l2 = ['a is greater', 'f is greater', 'c is greater']
# iterate over the elements of l2
for i in l2:
# check if the first letter of e is in l1
if i[0] not in l1:
print(i)

f is greater

您不需要在l2的元素上迭代两次，并且要检查值(i[0])是否在集合(l1)中使用in。

如果你想检查一个不同的位置，只需改变i的索引，例如，如果你想检查最后一个位置，执行:

l1 = ['a','b','c','d']
l2 = ['Greater is c', 'Greater is f', 'Greater is d']
# iterate over the elements of l2
for i in l2:
# check if the first letter of e is in l1
if i[-1] not in l1:  # note the -1
print(i)

Greater is f

如果您想要考虑在l1中不存在句子的所有单词(以空格分隔)，一种方法:

l1 = ['a', 'b', 'c', 'd']
l2 = ['Greater is c', 'Greater is f', 'f is greater', "Hello a cat"]
s1 = set(l1)
# iterate over the elements of l2
for i in l2:
# check if the first letter of e is in l1
if s1.isdisjoint(i.split()):
print(i) 

Greater is f
f is greater

如果检查字符串是否包含，请执行:

l1 = ['Book Date:', 'Statement Number:', 'Opening Book Balance:', 'Closing Book Balance:', 'Number Of Debits:',
'Number of Credits:', 'Total Debits:', 'Total Credits:', 'Report Date:', 'Created by:', 'Modified by:', 'Printed by:']
l2 = ['<p>Book Date: 06-01-21 To 06-30-21</p>', '<p>Statement Number: 126 </p>', '<p>this value need to print</p>']

# iterate over the elements of l2
for i in l2:
# check if the first letter of e is in l1
if not any(j in i for j in l1):
print(i)

<p>this value need to print</p>

Answer 2

给你:

ref = ('a','b','c','d')
l2 = ['a is greater', 'f is greater', 'c is greater']
l3 = ['Greater is c', 'Greater is f', 'Greater is d']
l4 = ['...c...', '...f...', '...d...']
[item for item in l2 if not item.startswith(ref)] # 'f is greater'
[item for item in l3 if not item.endswith(ref)]   # 'Greater is f'
[item for item in l4 if not any(letter in item for letter in ref)] # '...f...'