问题描述
如何遍历字符串中的每个第二个元素?
一种方法是(如果我想遍历第 n 个元素):
One way to do this would be(If I want to iterate over nth element):
sample = "This is a string"
n = 3 # I want to iterate over every third element
i = 1
for x in sample:
if i % n == 0:
# do something with x
else:
# do something else with x
i += 1
是否有任何pythonic"方式来做到这一点?(我很确定我的方法不好)
Is thery any "pythonic" way to do this? (I am pretty sure my method is not good)
推荐答案
如果你想每隔 n 步做一些事情，其他情况下做其他事情，你可以使用 enumerate 来获取索引，并使用模数:
If you want to do something every nth step, and something else for other cases, you could use enumerate to get the index, and use modulus:
sample = "This is a string"
n = 3 # I want to iterate over every third element
for i,x in enumerate(sample):
if i % n == 0:
print("do something with x "+x)
else:
print("do something else with x "+x)
注意它不是从 1 而是从 0 开始的.如果你想要别的东西，可以给 i 添加一个偏移量.
Note that it doesn't start at 1 but 0. Add an offset to i if you want something else.
要仅对每个第 n 个元素进行迭代，最好的方法是使用 itertools.islice 来避免创建硬"字符串来仅对其进行迭代:
To iterate on every nth element only, the best way is to use itertools.islice to avoid creating a "hard" string just to iterate on it:
import itertools
for s in itertools.islice(sample,None,None,n):
print(s)
结果:
T
s
s
r
g
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