如何进行累积的“全部"操作? [英] How to do a cumulative "all"

问题描述

设置
考虑numpy数组a

Setup
Consider the numpy array a

>>> np.random.seed([3,1415])
>>> a = np.random.choice([True, False], (4, 8))

>>> a
array([[ True, False,  True, False,  True,  True, False,  True],
       [False, False, False, False,  True, False, False,  True],
       [False,  True,  True,  True,  True,  True,  True,  True],
       [ True,  True,  True, False,  True, False, False, False]], dtype=bool)

问题
对于每一列，我都想确定所有列的累积等值.

Question
For each column, I want to determine the cumulative equivalent for all.

结果应如下所示:

array([[ True, False,  True, False,  True,  True, False,  True],
       [False, False, False, False,  True, False, False,  True],
       [False, False, False, False,  True, False, False,  True],
       [False, False, False, False,  True, False, False, False]], dtype=bool)

采用第一列

a[: 0]

# Original First Column
array([ True, False, False,  True], dtype=bool)
# So far so good
#        \     False from here on
#         |    /---------------\
array([ True, False, False, False], dtype=bool)
# Cumulative all

所以基本上，只要我们有True，然后从第一个False

So basically, cumulative all is True as long as we have True and turns False from then on at the first False

我尝试过的事情
我可以用

What I have tried
I can get the result with

a.cumprod(0).astype(bool)

但是，当我知道从我看到的第一个False开始都是False时，我不禁想知道是否需要执行每个乘法.

But, I can't help but wonder if its necessary to perform each and every multiplication when I know everything will be False from the first False I see.

考虑更大的一维数组

b = np.array(list('111111111110010101010101010101010101010011001010101010101')).astype(int).astype(bool)

我认为这两个问题会产生相同的答案

I contend that these two produce the same answer

bool(b.prod())

和

b.all()

但是b.all()可能会短路，而b.prod()不会.如果我给他们计时:

But b.all() can short circuit while b.prod() does not. If I time them:

%timeit bool(b.prod())
%timeit b.all()

100000 loops, best of 3: 2.05 µs per loop
1000000 loops, best of 3: 1.45 µs per loop

b.all()更快.这意味着我必须有一种方法来进行累积的所有操作，这要比我的a.cumprod(0).astype(bool)

b.all() is quicker. This implies that there must me a way to conduct a cumulative all that is quicker that my a.cumprod(0).astype(bool)

推荐答案

所有ufunc都有5种方法:reduce，accumulate，reduceat，outer和at.在这种情况下，请使用accumulate，因为它会返回ufunc的累积应用程序的结果:

All ufuncs have 5 methods: reduce, accumulate, reduceat, outer, and at. In this case, use accumulate since it returns the result of cumulative applications of the ufunc:

In [41]: np.logical_and.accumulate(a, axis=0)
Out[50]: 
array([[ True, False,  True, False,  True,  True, False,  True],
       [False, False, False, False,  True, False, False,  True],
       [False, False, False, False,  True, False, False,  True],
       [False, False, False, False,  True, False, False, False]], dtype=bool)

---

In [60]: np.random.seed([3,1415])

In [61]: a = np.random.choice([True, False], (400, 80))

In [57]: %timeit np.logical_and.accumulate(a, axis=0)
10000 loops, best of 3: 85.6 µs per loop

In [59]: %timeit a.cumprod(0).astype(bool)
10000 loops, best of 3: 138 µs per loop

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