打印错误子列表的索引？答案

【问题标题】：Print index of sublists with error?打印错误子列表的索引？
【发布时间】：2013-08-03 01:41:22
【问题描述】：

我有一个由一定长度的子列表组成的列表，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特定长度的子列表的索引，然后将该索引打印出来。例如，在示例列表中，最后一个子列表的长度不是四，所以我会打印list 2 does not have correct length。这是我目前所拥有的：

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！

【问题讨论】：

标签： list python-3.x indexing

【解决方案1】：

当您需要索引和对象时，通常可以使用enumerate，它会产生(index, element) 元组。例如：

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

等等：

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")

生产

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

【讨论】：

这真的很有帮助！谢谢！只是想知道，有没有办法没有对象，只有索引？
不过，您需要让对象取其长度。您可以使用for i in range(len(newlist)):，然后使用newlist[i]，但这被认为是非pythonic 风格。

【解决方案2】：

我认为 DSM 得到了您想要的答案，但您也可以使用 index method 并编写如下内容：

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))

【讨论】：