将所有连续的都压入python列表中 [英] squash all consecutive ones in a python list

问题描述

我有用 0s 和 1s 填充的变量列表，例如:

I have variable lists filled with 0s and 1s, like:

l1 = [1,1,1,0,1,1]
l2 = [0,1,1,0,1,1,0,0,1]

什么是创建新列表以压缩所有连续 1s 的最有效方法.

what is the most efficient way to create new lists that squash all consecutive 1s.

因此结果将是:

l1_new = [1,0,1]
l2_new = [0,1,0,1,0,0,1]

提示:numpy/向量化或某些逻辑运算会很棒！

Hint: numpy/vectorization or some logical operation would be great!

推荐答案

这是与 np.diff 和按位运算:

l1 = [1,1,1,0,1,1]
l2 = [0,1,1,0,1,1,0,0,1]

a = np.array(l2)
a[~((np.diff(a,prepend=False)==0) & (a==1))]
# array([0, 1, 0, 1, 0, 0, 1])

或者对于第一个示例:

a = np.array(l1)
a[~((np.diff(a,prepend=False)==0) & (a==1))]
#array([1, 0, 1])

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