蟒蛇：从字符串共性 [英] Python: Get common characters from strings

问题描述

所以，newbious问题：我在寻找比较两个字符串，并能够回来，作为单独字符串的最小（和最好）的方式：

So, a newbious question: I'm looking for the smallest (and nicest) way of comparing two strings and being able to get back, as separate strings:

• 所有常用的字符，


• 的罕见字符（所有字符但没有常见的）


• 人物所特有的一根弦。

...使用Python（或Perl，如果是比较容易，但是preferably的Python）。例如：

...using Python, (or Perl, if it's easier—but preferably Python). Example:

A = "123 ABC"
B = "135 AZ"

thingamajigger(A, B) would give all these:

intersect = "13 A" (inclues space)
exclusion = "2BCZ5"
a_minus_b = "2BC"
b_minus_a = "5Z"

a_minus_b很简单......但如果​​有这些花哨的单行方式之一，把它关闭，然后我打开。

a_minus_b is quite simple... but if there's one of those fancy one-liner ways to pull it off, then I'm open.

for i in B:
    A = A.replace(i, "")

这是一个有点像串布尔运算。

It's a bit like boolean operations on strings.

（和一个巨大的奖金的人谁可以找到一种方式，将返回所有常见/不常见的字符从任意数量的字符串作为输入的。这将是的确非常方便。）

(And a massive bonus to anyone who can find a way that will return all common/uncommon chars from any number of strings as input. That would be very handy indeed.)

无论如何，感谢大家！

推荐答案

使用 设置 ：

s = set("123 ABC")
t = set("135 AZ")
intersect = s & t # or s.intersection(t)
exclusion = s ^ t # or s.symmetric_difference(t)
a_minus_b = s - t # or s.difference(t)
b_minus_a = t - s # or t.difference(s)

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