Question

我必须创建一个列表列表代码，它返回＆＃34; True＆＃34;确切地说，当所有列表中存在（至少1个）和喜欢的乐队时（可能> 3）。我对编程非常陌生，我很遗憾该怎么做。我尝试了一些东西，但它没有完全填满。我需要一些帮助！

favoriteBandLists = [
    ["Metallica","Linkin Park","Alice In Chains","Nirvana", "Soundgarden"],
    ["Pink Floyd","Alice In Chains","Soundgarden","Metallica","Linkin Park"],
    ["Audioslave","Offspring","Soundgarden","Linkin Park","The Beatles"]]

我的代码：

def commonFavoriteBand(list):
    foundCounterExampleyet = False
    for i in range(()):
        if(()):
            foundCounterExampleYet = True
    return not(foundCounterExampleYet)

print (commonFavoriteBand())

def commonFavoriteBandA():
    foundExampleYet = False
    for value in values:
        if():
            foundExampleYet = True

return foundExampleYet

Answer 1

我会说最简单但最容易理解的方式是：

favorite_band_lists = [
    ["Metallica", "Linkin Park", "Alice In Chains", "Nirvana", "Soundgarden"],
    ["Pink Floyd", "Alice In Chains", "Soundgarden", "Metallica", "Linkin Park"],
    ["Audioslave", "Offspring", "Soundgarden", "Linkin Park", "The Beatles"],
]

def common_favorite_band(band_lists):
    # If there are no bands at all, you can't really say there are any in
    # common
    if not band_lists:
        return False

    # Convert the first band list to a "set" -- a group of unique items
    common_bands = set(band_lists[0])

    # Then, for every other band list...
    for bands in band_lists[1:]:
        # ...intersect it with the running set.  This means `common_bands`
        # should throw away anything that isn't also in `bands`.
        common_bands.intersection_update(bands)

    # Now common_bands contains only the bands that are in every list.
    # But you wanted True or False, so cast it to a bool -- an empty set
    # will become False, a set with at least one item will become True.
    return bool(common_bands)

print(common_favorite_band(favorite_band_lists))  # True!

（顺便说一句，函数和变量名在Python中传统上用snake_case编写，而不是camelCase）

Answer 2

您可以使用：

s=set(favorite_band_lists[0])
for ii in favorite_band_lists[1:]:
    s=s.intersection(s,ii)
    if len(s) == 0:
        break

一旦交叉路口为空，这将停止，这可能是理想的。它还返回列表中通用的波段（如果有）。要生成True或False，请检查s的长度 - 交叉点列表 - 是否为＆gt; 0

print len(s) > 0

Answer 3

这是与其他人不同的解决方案：

def common_favorite_band(band_lists):
    for band in band_lists[0]:
        count = 0
        for band_list in band_lists[1:]:
            if band  not in band_list:
                break
            else:
                count += 1
        if count == len(band_lists) - 1:
            return True
    return False

它查看第一个列表中的所有波段，并检查该波段是否在每个其他列表中。我们每次添加1来计数。对于任何波段，如果count达到band_lists的长度 - 1，那么我们可以返回True并完成。如果乐队在任何时候没有出现在另一个列表中，我们就会从循环的迭代中断开并尝试另一个乐队。如果我们检查了每个波段，并且我们的计数没有等于band_list - 1的长度，那么我们返回False。

Python-列表列表

3 个答案: