在 Python 中排除 collections.Counter 中的零

Question

有没有办法 collections.Counter 不 count/ignores 给定值（这里是 0）：

from collections import Counter
import numpy as np

idx = np.random.randint(4, size=(100,100))
most_common = np.zeros(100)
num_most_common = np.zeros(100)

for i in range(100):
    most_common[i], num_most_common[i] = Counter(idx[i, :]).most_common(1)[0]

所以如果 0 是最常见的值，它应该给出第二个最常见的值。另外，有没有办法避免这种情况下的for循环？

Answer 1

您可以执行以下操作，使用生成器仅对不为 0 的内容进行计数。

most_common = np.array([Counter(x for x in r if x).most_common(1)[0][0] for r in idx])
num_most_common = np.array([Counter(x for x in r if x).most_common(1)[0][1] for r in idx])

甚至

count = np.array([Counter(x for x in r if x).most_common(1)[0] for r in idx])
most_common = count[:,0]
num_most_common = count[:,1]

Answer 2

对于正数，我们可以使用 -

#  @Divakar
def bincount2D_vectorized(a):    
    N = a.max()+1
    a_offs = a + np.arange(a.shape[0])[:,None]*N
    return np.bincount(a_offs.ravel(), minlength=a.shape[0]*N).reshape(-1,N)

# Get binned counts per row, with each number representing a bin 
c = bincount2D_vectorized(idx)

# Skip the first element, as that represents counts for 0s.
# Get most common element and count per row
most_common = c[:,1:].argmax(1)+1
num_most_common = c[:,1:].max(1)
# faster : num_most_common = c[np.arange(len(most_common)),most_common]

对于通用的 int 数字，我们可以像这样扩展 -

s = idx.min()
c = bincount2D_vectorized(idx-s)
c[:,-s] = 0
most_common = c.argmax(1)
num_most_common = c[np.arange(len(most_common)),most_common]
most_common += s

在 Python 中排除 collections.Counter 中的零

Exclude zeros in collections.Counter in Python

python

performance

counter

numpy

vectorization