排除集合中的零.Python中的计数器

Question

有没有办法让collections.Counter不计算/忽略给定的值(这里为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]

因此，如果0是最常见的值，它应该给出第二个最常见的值。此外，在这种情况下有没有办法避免 for 循环？

Answer 1

对于正数，我们可以使用vectorized-bincount - bincount2D_vectorized-

# https://stackoverflow.com/a/46256361/ @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

Answer 2

您可以执行以下操作，使用生成器仅在不为 0 时对其进行计数。

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]