Question

给出清单A：

#include <stdio.h>
#include <string.h>

int main(void)
{
    char line[100];
    size_t size;

    size = (size_t)printf("Enter text: ");
    fgets(line, 100, stdin);
    printf("\033[A\033[%zuC", size + strlen(line));
    fputs("   <-- What a great text!", stdout);
    return 0;
}

并列出B：

['cheese', 'spam', 'sausage', 'eggs']

我想从列表B中获取包含列表A中单词的字符串的代码。

即。对于像['cz', 'sp', 'sg', 'eg']这样的字符串，我想从列表B中获取代码something about cheese。保证输入字符串中只出现列表A中的一个项目。

如何在不检查每种情况的情况下实现这一目标？即什么是更好的方式，而不是：

cz

Answer 1

zip他们在一起并迭代A，直到一个人在字符串中。然后返回相应的B.

def foo(someString, listA, listB):
    for a,b in zip(listA, listB):
        if a in someString:
            return b

someString = "something about cheese"
listA = ['cheese', 'spam', 'sausage', 'eggs']
listB = ['cz', 'sp', 'sg', 'eg']
print(foo(someString, listA, listB))  # => cz

Answer 2

mapping = {
    'cheese': 'cz',
    'spam': 'sp',
    'sausage': 'sg',
    'eggs': 'eg'
}

s = 'something about cheese'

for key, value in mapping.iteritems():
    if key in s:
        return value

Answer 3

为什么不使用这样的东西：

for i, key in enumerate(list_a):
    if key in s:
        return list_b[i]

或

for key, value in zip(list_a, list_b):
    if key in s:
        return value

Answer 4

List_A = ['cheese', 'spam', 'sausage', 'eggs']
List_B = ['cz', 'sp', 'sg', 'eg']

s = 'something about cheese'

SplitWordsList = s.split ( " " ) # Separate the words in the sentence

for SplitWord in SplitWordsList :
    if  ( SplitWord in List_A ) :
        return  ( List_B [ List_A.index ( SplitWord ) ] )
        break

Python部分字符串映射到列表中的值

4 个答案: