如何计算 numpy 数组中 TRUE 到 FALSE 的数量？

Question

我正在尝试计算 numpy.array 中 True 值的数量，直到达到 False。

例如，如果您有

condition = np.array([True, True, True, False, 
    False, True, False, True, True, False, True])

那么想要的结果就是

np.array([3, 2, 1, 0, 0, 1, 0, 2, 1, 0, 1])

编辑：

Numpy first occurrence of value greater than existing value is not what I'm asking because that's asking about the first time a condition is satisfied. I'm asking how many times in a row a condition is satisfied each time it is satisfied. Find first sequence item that matches a criterion 也不起作用，因为我再次询问的不是序列第一次满足条件的情况。

Answer 1

condition = np.array([True, True, True, False, 
    False, True, False, True, True, False, True])
    
r = []
c = 0

for x in reversed(condition):
  if x == False:
    c = 0
  else:
    c+=1
  r.append(c)
  
r.reverse()
print(np.array(r))

https://trinket.io/python3/a5bd54189b

一个较短的版本：

r = [0]
for x in reversed(condition):
  r.append(r[-1] + 1 if x else 0)

r.reverse()
r.pop()
print(np.array(r))

Answer 2

import itertools
condition = np.array([True, True, True, False, False, True, False, True, True, False, True])
group_list = [list(g) for _, g in itertools.groupby(enumerate(condition), key = lambda x: x[-1])]
temp = 0
ans = []
for item in group_list:
    if item[0][1]:
        for i in range(len(item)):
            ans.append(len(item) - (item[i][0] - temp))
        temp += len(item)
    else:
        for i in range(len(item)):
            ans.append(0)
        temp += len(item)
print(ans)  
>>> [3, 2, 1, 0, 0, 1, 0, 2, 1, 0, 1]

Answer 3

你实际上可以用两行代码完全做到这一点，而且没有扩展库：

false_indices = [np.where(condition[j:] == False)[0] for j in range(len(condition))]

result = np.array([arr[0] if arr.size != 0 else int(condition[-1]) for arr in result])

如果不是 conditions 中的最后一个 bool 会引发 IndexError.

，它甚至可能是单行代码

Answer 4

逐行解释计算

import numpy as np                                                      

condition = np.array([True, True, True, False, False, True, False, True, True, False, True])                       

rvs = condition[::-1]   # Reverse the array order
rvs                                                                     
# array([ True, False,  True,  True, False,  True, False, False,  True,
        True,  True])

cum_rvs = rvs.cumsum()  # Cumulative sum of Trues                                         
cum_at_zero = (~rvs)*cum_rvs   # Cum where rvs is True set to zero                                         

cum_max = np.maximum.accumulate( cum_at_zero ) # Equivalent to cummax( cum_at_zero )
cum_max                                                               
# array([0, 1, 1, 1, 3, 3, 4, 4, 4, 4, 4])
  
result = cum_rvs - cum_max  # Use cum_max to adjust the cum_rvs down                        

result = result[::-1]  # Reverse the result order                                                
result                                                                  
# array([3, 2, 1, 0, 0, 1, 0, 2, 1, 0, 1])

或更简洁的形式：

rvs = condition[::-1]                                                   
cum_rvs = rvs.cumsum()

result = ( cum_rvs - np.maximum.accumulate((~rvs)*cum_rvs ))[::-1]

如何计算 numpy 数组中 TRUE 到 FALSE 的数量？

How to count the number of TRUE until a FALSE in numpy array?

python

boolean

numpy

counting