仅删除 alpha 重复项

Question

在 Python 中，我想从字符串中删除重复的字母，而不是数字或空格。我想出了：

result = []
seen = set()
for char in string:
    if char not in seen:
        seen.add(char)
        result.append(char)
return "".join(result)

但这使得：

>>> delete_duplicate_letters("13 men were wounded in an explosion yesterday around 3:00pm.")
13 menwroudiaxplsyt:0.

当我想要时：

>>> delete_duplicate_letters("13 men were wounded in an explosion yesterday around 3:00pm.")
13 men wr oud i a xpls yt 3:00.

我尝试使用letter 代替char、isalpha() 函数和if int 语句等，但我什么都做不了。

Answer 1

>>> from string import digits, whitespace
>>> from collections import OrderedDict
>>> s = set(whitespace + digits)
>>> ''.join(OrderedDict((object() if c in s else c, c) for c in text).values())
'12 men wr oud i a xpls yt  3:00.'

object() 此处仅用于确保您要离开的字符的键始终是唯一的，因为object() 每次都会创建不同的对象。其他字符本身用作键，因此过滤掉重复项。

Answer 2

试试这个：

result = ""
for char in string:
    if not (char.isalpha() and char in result):
        result += char

Answer 3

使用str.isspace 和str.isdigit：

strs = "13 men were wounded in an explosion yesterday around 3:00pm."
result = []
seen = set()
for char in strs:
    if char not in seen:
        if not (char.isspace() or char.isdigit()):
           seen.add(char)
        result.append(char)
print "".join(result)

输出：

13 men wr oud i a xpls yt  3:00.

Answer 4

看来你快到了。您可以在循环中添加一些检查：

result = []
seen = set()
for char in string:
    if char.isdigit() or char.isspace():
        result.append(char)
    elif char not in seen:
        seen.add(char)
        result.append(char)
return "".join(result)