Question

我有两个清单：

list1 = ['home', 'school', 'bus', football']
list2 = ['yesterday I went to home', 'I am busy', 385723, 'I feel like 
         playing football', I was tired last week', 'I go to school by 
         bus',' there is a bus stop near my home', 'I am hungry']

我想知道如何打印list2中包含list1中任何项目（至少1个）的任何项目？例如，在我们的示例中，应打印以下内容：

'yesterday I went to home','I feel like playing football', 'I go to school 
 by bus',' there is a bus stop near my home'

我写了一段代码，但运行时我的应用程序中断了：

    theList = []
    i = 0
    while i < len(list1):
        for element in list2:
            if (list1[i]) in element.lower():
                theList.append(element)
        i += 1
    print(errorList)

Answer 1

另一种解决方案：

print([el2 for el2 in list2 if any(el1 in str(el2).split() for el1 in list1)])

Answer 2

试试这个：

for string in list2:
    if(type(string)==str): #if the element in list2 is a string
        if(set(string.split()) & set(list1)): #split the string into words, and check if it has a set intersection with list1
            print(string)

输出：

yesterday I went to home
I feel like playing football
I go to school by bus
 there is a bus stop near my home

Answer 3

试试这个！它来自记忆，但它应该工作：）

# Print Your code here
print('Hello in console window')

list1 = ['home', 'school', 'bus', 'football']
list2 = ['yesterday I went to home', 'I am busy', 385723,
         'I feel like playing football', 'I was tired last week'
         , 'I go to school by bus'
         ,' there is a bus stop near my home', 'I am hungry']

theList = []
i = 0
for item in list2:
    item1 = str(item).split(" ")
    if len(item1) >= 1:
        is_in = [x for x in item1 if x.lower() in list1]
    else:
        is_in = item if item.lower() == list1 else 0
    if len(is_in) > 0:
        theList.append(item)

for item in theList:
    print(item)

让我知道它是否错了，

Answer 4

需要分成单词并匹配。

list1 = ['home', 'school', 'bus', 'football']
list2 = ['yesterday I went to home', 'I am busy', 385723, 'I feel like playing football',
         'I was tired last week', 'I go to school by bus',' there is a bus stop near my home', 'I am hungry']

theList = []
for item2 in list2:
    for item1 in list1:
        bfound=False
        if type(item2) is  str:
            words = item2.lower().split()
            for word in words:
                if word==item1:
                    theList.append(item2)
                    bfound=True
                    break
            if bfound:
                break

print(theList)

输出

['yesterday I went to home', 'I feel like playing football', 'I go to school by bus', ' there is a bus stop near my home']

Answer 5

list1 = ['home', 'school', 'bus', 'football']
list2 = ['yesterday I went to home', 'I am busy', '385723', 
        'I feel like playing football', 'I was tired last week', 
        'I go to school by bus',' there is a bus stop near my home', 
        'I am hungry']

ans = []
for s2 in list2:
    l2 = s2.split(' ')
    for ss in l2:
        if ss in list1:
           ans.append(s2)
           break
print ans

Answer 6

这是一个易于阅读的多行解决方案，它没有考虑list2类型不属于string的项目：

result = []
for sentence in list2:
    if isinstance(sentence, str):
        for word in list1:
            if word in sentence.split():
                result.append(sentence)
                break
print(result)

这里也是一个单行风格的解决方案（我看到它已经由@grimek提供），它将list2中的任何项目转换为string类型：

result = [sentence for sentence in list2 if any(word in str(sentence).split() for word in list1)]
print(result)

Answer 7

list3 = [s for s in list2 if (type(s)==str) and (set(s.split())&set(list1))]

试试这个： - ）

list3的结果是

<'>''昨天我回家'，'我觉得自己喜欢踢足球'，'我去了乘公共汽车上学'，'我家附近有一个公共汽车站']

Answer 8

如果您直接检查是否有“公共汽车”，那就有一个问题。在string中，你可能会得到错误的结果，因为它会检查sub_string by_sub_string，所以如果你在这个例子中这样做：

'bus' in 'I am busy' then it will return True :

所以解决方案不是检查sub_string，而是逐字检查：

print([j for i in list1 for j in list2 if isinstance(j,str) if i in j.split()])

输出：

['yesterday I went to home', ' there is a bus stop near my home', 'I go to school by bus', 'I go to school by bus', ' there is a bus stop near my home', 'I feel like playing football']

另一个列表中列表中的项目

8 个答案: