IEnumerable:在与谓词匹配的最后一个字符之前获取所有字符 [英] IEnumerable: Get all before the last that matches a predicate

问题描述

我有这样的IEnumerable<int>，只是更长的时间:

I have an IEnumerable<int> like this, only longer:

5, 0, 0, 0, 0, 4, 0, 0, 0, 2, 0, 6, 0, 0, 0, 0, 0

现在我想返回最后一个非零值之前的所有元素:

and now I want to return all elements before the last non-zero value:

5, 0, 0, 0, 0, 4, 0, 0, 0, 2, 0

sequence.Last()似乎无济于事，因为它返回的是最后一次出现的索引，而不是最后一次出现的索引.

It seems sequence.Last() doesn't help here, because it returns the last occurrence, not the index of the last occurrence.

我曾考虑使用

var lastIndex = sequence.LastIndex(x=>x!=0);
subsequence = sequence.Take(lastIndex);

在通常情况下可以使用，但LastIndex不存在，或者

which would work in the general case, but LastIndex doesn't exist, or

var last = sequence.Last(y=>y!=0);
subsequence = sequence.TakeWhile(x=>x!=last)

在示例中有效，但在通常情况下(可能存在重复的非零值)无效.

which would work on the example, but not in the general case, where there may be duplicated non-zero values.

有什么想法吗?

推荐答案

您可以尝试

var allDataBeforeLastNonZero= sequence.GetRange(0,sequence.FindLastIndex(x=>x!=0));

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