Question

我怎样才能检查一个列表中的单词是否等于另一个列表中的单词？例如，我有三个列表：

["fish", "boat", "oar"], ["rod", "gunwale", "fish", "net"], ["net", "hook", "weight"]

如何检查第一个列表中的单词是否出现在任何其他列表中？例如，我如何迭代其他两个列表中的每个单词以查看单词＆＃34; fish＆＃34;在他们身上，和＃34;船一样＃34;并且＆＃34;桨。＆＃34;

Answer 1

您可以使用in运算符：

l1 = ["fish", "boat", "oar"]
l2 = ["rod", "gunwale", "fish", "net"]
l3 = ["net", "hook", "weight"] 

for w in l1:
  if w in l2:
    print 'found %s in l2!' % w
  if w in l3:
    print 'found %s in l3!' % w

如果你想检查它是否在其他两个列表的任何一个中，你可以组合它们并在里面进行相同的检查：

if w in l2 + l3:
  print 'found %s in another list!'

Demo

Answer 2

您可以使用set intersection功能，例如：

s1 = set(["fish", "boat", "oar"]) 
s2 = set(["rod", "gunwale", "fish", "net"])
s3 = set(["net", "hook", "weight"])
commonS12 = s1 & s2#gives you common elements

Answer 3

根据您的最新评论，您似乎希望计算包含第一个列表元素的列表数量。这是一个小功能，它正是这样做的：

def count_in_lists(e, lol):
    """
    Count how many of the lists passed in the list of lists 'lol' contain
    the element 'e'.
    """
    count = 0
    for current_list in lol:
        if e in current_list:
            count += 1
    return count

现在，类似于casraf的回答，遍历l1，并使用当前元素count_in_lists()作为第一个参数调用函数l1，并列出包含所有其他列表的列表'作为第二个参数感兴趣：

for w in l1:
    print("'{}' is contained in {} other lists".format(
                w, 
                count_in_lists(w, [l2, l3])))

这为您提供了此输出：

'fish' is contained in 1 other lists
'boat' is contained in 0 other lists
'oar' is contained in 0 other lists

Answer 4

l1 = ["fish", "boat", "oar"]
l2 = ["rod", "gunwale", "fish", "net"]
l3 = ["net", "hook", "weight"]

print("Intersection between l1 and l2",list(set(l1).intersection(l2)))
print("Intersection between l2 and l3",list(set(l2).intersection(l3)))
print("Intersection between l1 and l3",list(set(l1).intersection(l3)))

print("Intersection between l1 and l2 and l3",list(set(l1).intersection(l2).intersection(l3)))

Answer 5

您可以在一个列表中收集所有三个列表：

lists = [["fish", "boat", "oar"], ["rod", "gunwale", "fish", "net"], ["net", "hook", "weight"]]

然后遍历列表列表，并对当前列表中的每个元素使用in运算符检查它是否在任何其他列表中：

import itertools
for lst in lists:
    # this chains all lists different than the current one into one flat list
    other_lists = list(itertools.chain(*filter(lambda l: l is not lst, lists)))
    for elm in lst:
        if elm in other_lists:
            print '%s found in another list' % elm

这是一个通用答案，适用于任意数量的列表，而不仅仅是三个。

Answer 6

您也可以使用列表理解来完成此操作。

l1 = ["fish", "boat", "oar"]
l2 = ["rod", "gunwale", "fish", "net"]
l3 = ["net", "hook", "weight"] 

print(["{0} is in {1}".format(x, [m for m, li in globals().items() if li is L]) for x in l1 for L in [l2,l3] if x in L][0])
# use globals().iteritems() in python 2

输出：

"fish is in [l1]"

Answer 7

如果数据与OP建议的结构不同。

question_data = ques_data = [["fish", "boat", "oar"], ["rod", "gunwale", "fish", "net"], ["net", "hook", "weight"]]

我们可能需要采取不同的方法。此外，在for循环中执行此操作非常重要，因此无论列表有多长，它都可以进行搜索：

l1_values = ques_data[0]
remaining_list = ques_data[1::]

print(l1_values) # gives
['fish', 'boat', 'oar']

print(remaining_list) # gives
[['rod', 'gunwale', 'fish', 'net'], ['net', 'hook', 'weight']]

count = 1
for x in l1_values:
    for lists in remaining_list:
        count =+ 2
        if x in lists:
        print(x + ' is in list #' + str(count))

# gives the answer as:
fish is in list #2

        print(x + ' is in list ' + str(lists))

# gives the answer as:
fish is in list ['rod', 'gunwale', 'fish', 'net']

将一个列表中的单词与其他列表中的单词进行比较

7 个答案: