Question

我有一个列表，该列表由一定长度的子列表组成，有点像[["a","b","c","d"],["a","b","c","d"],["a","b","c","d","e"]]。我想要做的是找到没有那个特定长度的子列表的索引，然后打印出该索引。例如，在示例列表中，最后一个子列表的长度不是四个，因此我将打印list 2 does not have correct length。这就是我到目前为止所做的：

for i in newlist:
    if len(i) == 4:
        print("okay")
elif len(i) != 4:
    ind = i[0:]     #this isnt finished; this prints out the lists with the not correct length, but not their indecies 
    print("not okay",ind)

提前致谢！

Answer 1

如果同时需要索引和对象，通常可以使用enumerate，这会产生(index, element)个元组。例如：

>>> seq = "a", "b", "c"
>>> enumerate(seq)
<enumerate object at 0x102714eb0>
>>> list(enumerate(seq))
[(0, 'a'), (1, 'b'), (2, 'c')]

所以：

newlist = [["a","b","c","d"],["a","b","c","d"],["a","b","c","d","e"]]

for i, sublist in enumerate(newlist):
    if len(sublist) == 4:
        print("sublist #", i, "is okay")
    else:
        print("sublist #", i, "is not okay")

产生

sublist # 0 is okay
sublist # 1 is okay
sublist # 2 is not okay

Answer 2

我认为DSM得到了你所追求的答案，但你也可以使用index method并写下如下内容：

new_list = [["a","b","c","d"],["a","b","c","d"],["a","b","c","d","e"]]
size_filter = 4
# all values that are not 4 in size
indexes = [new_list.index(val) for val in new_list if len(val)!=size_filter]
# output values that match
for index in indexes:
    print("{0} is not the correct length".format(index))

打印有错误的子列表索引？

2 个答案: