Python-计算符号变化 [英] Python - counting sign changes

问题描述

我有一个从左到右阅读的数字列表.每当我在阅读要计数的序列时遇到符号变化时.

I have a list of numbers I am reading left to right. Anytime I encounter a sign change when reading the sequence I want to count it.

X = [-3,2,7,-4,1,-1,1,6,-1,0,-2,1] 
X = [-, +, +, -, +, -, +, +, -, -,-,+]

因此，此列表中有8个符号更改.

So, in this list there are 8 sign changes.

如果项目[0](在这种情况下为-3)为负，则视为符号更改.同样，列表中的任何0都被视为[-].

When Item [0] (in this case -3) is negative it is considered a sign change. Also, any 0 in the list is considered [-].

任何帮助将不胜感激.

Any help would be greatly appreciated.

推荐答案

您可以使用 itertools.groupby 计算正数和非正数的组:

You can use itertools.groupby to count the groups of positive and non-positive numbers:

>>> x = [-3,2,7,-4,1,-1,1,6,-1,0,-2,1] 

>>> import itertools
>>> len(list(itertools.groupby(x, lambda x: x > 0)))

结果:

8

在您的问题中，您要求:

In your question you state that you want:

• 计算更改，而不是分组
• 如果第一个元素不是正数，则计算额外的变化.

您可以通过直接测试第一个元素并调整结果来做到这一点:

You can do this either by testing the first element directly and adjusting the result:

>>> len(list(itertools.groupby(x, lambda x: x > 0))) - (x[0] > 0)

或在进行分组之前在输入之前加上一个正数，然后从结果中减去1:

or by prepending a positive number to the input before doing the grouping then subtracting 1 from the result:

>>> len(list(itertools.groupby(itertools.chain([1], x), lambda x: x > 0))) - 1

当心您的输入列表是否可以为空-前一种解决方案将引发异常.

Watch out if your input list could by empty - the former solution will raise an exception.

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