Question

我想计算列表项满足一个条件的次数。例如a [i]＆gt; 1：

a = [2,4,3,0]

counter = 0
for value in a:
    if value > 1:
        counter += 1

是否可以使用enumerate函数来避免计数器+ = 1？

正如在循环中所做的那样：

for i,j in enumerate(list(xrange(5))):
    print i
    print j

Answer 1

你可以使用列表理解 - 奖励积分，因为它们很快（列表补偿用C实现）

a = [2,3,4,0]
count = len([i for i in a if i > 1])
# Or, to avoid a temporary list: (courtesy of John Kugelman)
count = sum(1 for i in a if i > 1)

说明：

len(...) # Gives the number of terms in the list
[i for i in a ... ] # Works like a for loop- this list is composed of pieces named i, where
                    # i is each term in a
[ ... if i > 1] # As long as that i is > 1.

# The sum() method does the same thing, but slightly more memory-efficient

可以枚举用于计算某些条件的发生吗？

1 个答案: