交替换弦 [英] alternating string replace

问题描述

说我有一个如下字符串：

s1 =''hi_cat_bye_dog''

和我想用''：''替换偶数''_''和用'''''取消偶数''_'''，'

以便我得到一个如下所示的新字符串：

s2 =''嗨：猫，再见：狗''

有没有共同的方法可以做到这一点？我不能提出任何

解决方案...


提前谢谢

Cesco

解决方案

1月9日上午11：34，cesco< fd.calabr ... @ gmail.comwrote：

说我有一个如下字符串：

s1 =''hi_cat_bye_dog''

我希望用'''''替换偶数''_''和用'''，'

替换奇怪''_''以便我得到一个像以下：

s2 =''嗨：猫，再见：狗''

有没有共同的方法可以实现这一目标？我不能提出任何

解决方案...


提前致谢

Cesco

为了替换其他每一个，我倾向于使用模数，使用

计数器。


count = （如果你想要以0开头，那么就做1）


/更换/在lalastring中：

如果计数％2 == 0：

甚至做

否则：

做奇数

算+ = 1


这是一种相当基本的方式，我相信人们会确保更有效的方式，但它通常适合我。


虽然我现在想知道当count == 2时是否计数％2是真的还是

false因为它返回0


当然你还需要做你的替换但你应该能够这样做。

cesco写道：

我想用''：''替换偶数''''和奇数''_''与''，''

以便我得到如下新字符串：

s2 =''hi：cat，bye：dog' '

有没有共同的方法可以实现这一目标？我不能提出任何

解决方案...

如何拆分_，加入对与' ;："，最后加入

结果与， ？

>> s1 =" hi_cat_bye_dog"
s1 = s1 .split（" _"）
s1

[''hi''，''cat''， ''bye''，''dog'']

>> s1 = [s1 [i] +"：" + s1 [i + 1] for i in range（0，len（s1），2）]
s1

[''hi：cat''，''bye：dog'']

< blockquote class =post_quotes>
>> s1 ="，" .join（s1）
s1

''嗨：猫，再见：狗''


（还有很多其他方法可以做到这一点，但上面的3线是短的，并且

直截了当。注意使用ra nge（）跳过其他所有项目

在列表中）


< / F>

cesco写道：

说我有一个如下字符串：s1 =''hi_cat_bye_dog''

我想用'''''替换偶数''_''和用''代替奇怪''_''''''

我得到一个像下面这样的新字符串：s2 =''hi：cat，bye：dog''

>>导入来自itertools导入周期
re.sub（" _"，lambda m，c = cycle（"，"）。next：c（），" hi_cat_bye_dog"）

''hi：cat，bye：dog''

有没有共同的方法来实现这一目标？我不能提出任何

解决方案...

有很多。如果你想学习Python，不要害怕以啰嗦的方式（带循环和辅助函数）来写它

。


彼得

Hi,

say I have a string like the following:
s1 = ''hi_cat_bye_dog''
and I want to replace the even ''_'' with '':'' and the odd ''_'' with '',''
so that I get a new string like the following:
s2 = ''hi:cat,bye:dog''
Is there a common recipe to accomplish that? I can''t come up with any
solution...

Thanks in advance
Cesco

解决方案

On Jan 9, 11:34 am, cesco <[e-mail]:

Hi,

say I have a string like the following:
s1 = ''hi_cat_bye_dog''
and I want to replace the even ''_'' with '':'' and the odd ''_'' with '',''
so that I get a new string like the following:
s2 = ''hi:cat,bye:dog''
Is there a common recipe to accomplish that? I can''t come up with any
solution...

Thanks in advance
Cesco

For replacing every other one, I tend to use the modulo, with a
counter.

count = (if u want even start with 0 else do 1)

while/for replacing /in lalastring:
if count % 2 == 0:
do even
else:
do odd
count += 1

This is a rather basic way of doing it, and I am sure people will sure
much more efficient ways, but it generally works for me.

Though I am now wondering if count % 2 when count == 2 is true or
false as it returns 0

Of course you still need to do your replace but you should be able to
do that.

cesco wrote:

and I want to replace the even ''_'' with '':'' and the odd ''_'' with '',''
so that I get a new string like the following:
s2 = ''hi:cat,bye:dog''
Is there a common recipe to accomplish that? I can''t come up with any
solution...

how about splitting on "_", joining pairs with ":", and finally joining
the result with "," ?

>>s1 = "hi_cat_bye_dog"
s1 = s1.split("_")
s1

[''hi'', ''cat'', ''bye'', ''dog'']

>>s1 = [s1[i]+":"+s1[i+1] for i in range(0,len(s1),2)]
s1

[''hi:cat'', ''bye:dog'']

>>s1 = ",".join(s1)
s1

''hi:cat,bye:dog''

(there are many other ways to do it, but the above 3-liner is short and
straightforward. note the use of range() to step over every other item
in the list)

</F>

cesco wrote:

say I have a string like the following: s1 = ''hi_cat_bye_dog''
and I want to replace the even ''_'' with '':'' and the odd ''_'' with '','' so
that I get a new string like the following: s2 = ''hi:cat,bye:dog''

>>import re
from itertools import cycle
re.sub("_", lambda m, c=cycle(":,").next: c(), "hi_cat_bye_dog")

''hi:cat,bye:dog''

Is there a common recipe to accomplish that? I can''t come up with any
solution...

There are many. If you want to learn Python don''t be afraid to write it
in a long-winded way (with loops and helper functions) first.

Peter

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