如何打破for循环？ [英] how to break a for loop?

问题描述

你好！

现在是圣彼得堡的1:56点，我还在编码...也许

'这就是我遇到愚蠢问题的原因：我需要从列表开头的

中删除零，但我不能:-(。我用


for i，coef in enumerate（coefs）：

if coef == 0：

del coefs [i]

else：

打破


但它会从列表中删除所有零。这是什么？

PS我觉得这个问题很愚蠢。 .. ;-)

Hello!
It''s 1:56 o''clock in St.-Petersburg now, and I am still coding... maybe
that''s why I encountered stupid problem: I need to remove zeros from
the begining of list, but I can''t :-(. I use

for i,coef in enumerate(coefs):
if coef == 0:
del coefs[i]
else:
break

but it removes ALL zeros from list. What''s up?
P.S. I feel SO stupid asking this quastion... ;-)

推荐答案

Gregory Petrosyan写道：

Gregory Petrosyan wrote:

你好！
它'现在是在圣彼得堡的1:56点，我还在编码...也许
这就是为什么我遇到了愚蠢的问题：我需要从
删除零列表的开头，但我不能:-(。我使用

for i，coef in enumerate（coefs）：
if coef == 0：
del coefs [i]
否则：
打破

但它会从列表中删除所有零。这是怎么回事？


我不知道枚举是如何实现的，但是我会怀疑在b循环中修改列表对象

到del要求麻烦


尝试

for i，coef in enumerate（coefs [：]）：

代替

PS我觉得愚蠢地问这个问题......; - ）

Hello!
It''s 1:56 o''clock in St.-Petersburg now, and I am still coding... maybe
that''s why I encountered stupid problem: I need to remove zeros from
the begining of list, but I can''t :-(. I use

for i,coef in enumerate(coefs):
if coef == 0:
del coefs[i]
else:
break

but it removes ALL zeros from list. What''s up?
I don''t know how enumerate is implemented, but I would
suspect that modifying the list object in the loop
through del is asking for trouble

try
for i,coef in enumerate(coefs[:]):
instead
P.S. I feel SO stupid asking this quastion... ;-)

uda4i


hth，Daniel

uda4i

hth, Daniel

zero_list = [0,0,0,0,0,1,2,3,4]

范围内的x（len（ zero_list））：

if zero_list [x]！= 0：break

nonzero_list = zero_list [x：]

print nonzero_list


但更多的是pythonic解决方案将从其他用户发布我

猜测:)


HTH

Petr Jakes

zero_list=[0,0,0,0,0,1,2,3,4]
for x in range(len(zero_list)):
if zero_list[x]!=0: break
nonzero_list=zero_list[x:]
print nonzero_list

but some more "pythonic" solutions will be posted from other users I
guess :)

HTH
Petr Jakes

我只花了几秒钟思考它:)


lst = [0,0,0,1,2,3,0 ，11]


尝试：

del lst [0：lst.index（0）]

除了ValueError：

通过


是更好的解决方案

I just spend some more seconds thinking about it :)

lst = [0,0,0,1,2,3,0,11]

try:
del lst[0:lst.index(0)]
except ValueError:
pass

is better solution

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