Question

我有一种感觉，我会被告知去初学者指南＆＃39;或者你有什么，但我在这里有这个代码

does = ['my','mother','told','me','to','choose','the']
it = ['my','mother','told','me','to','choose','the']
work = []

while 5 > len(work):
    for nope in it:
        if nope in does:
            work.append(nope)

print (work)

我得到了

['my', 'mother', 'told', 'me', 'to', 'choose', 'the']

这是为什么？我如何说服它返回

['my', 'mother', 'told', 'me']

Answer 1

您可以尝试这样的事情：

ABC  XUAS  BSNMM
CVD  nbvn  nbmsb
SVDB  NBV  KJHA
TTS  MNMN  NBA

您的代码存在的问题是，在完成for nope in it: if len(work) < 5 and nope in does: work.append(nope) else: break的所有项目并添加it中的所有项目后，它会检查工作的长度。

Answer 2

你可以这样做：

does = ['my','mother','told','me','to','choose','the']
it = ['my','mother','told','me','to','choose','the']
work = []
for nope in it:
    if nope in does:
        work.append(nope)
work = work[:4]
print (work)

它只是在不检查长度的情况下制作列表，然后将其剪切并仅留下4个第一元素。

Answer 3

或者，为了更接近原始逻辑：

i = 0
while 4 > len(work) and i < len(it):
    nope = it[i]
    if nope in does:
        work.append(nope)
    i += 1

# ['my', 'mother', 'told', 'me', 'to']

Answer 4

只是为了好玩，这里是一个没有进口的单线：

does = ['my', 'mother', 'told', 'me', 'to', 'choose', 'the']
it = ['my', 'mother', 'told', 'me', 'to', 'choose', 'the']
work = [match for match, _ in zip((nope for nope in does if nope in it), range(4))]

对于第三个列表中给定数量的元素，返回两个列表之间的字符串匹配

4 个答案: