用 pandas 替换字符串中字符的所有但最后一次出现的字符 [英] Replace all but last occurrences of a character in a string with pandas

问题描述

使用Pandas删除字符串中除最后一个期间外的所有内容，如下所示:

using Pandas to remove all but last period in a string like so:

s = pd.Series(['1.234.5','123.5','2.345.6','678.9'])
counts = s.str.count('\.')
target = counts==2
target
0     True
1    False
2     True
3    False
dtype: bool

s = s[target].str.replace('\.','',1)
s
0    1234.5
2    2345.6
dtype: object

但是，我想要的输出是:

my desired output, however, is:

0    1234.5
1    123.5
2    2345.6
3    678.9
dtype: object

replace命令和mask目标似乎正在丢弃未替换的值，我看不出如何解决此问题.

The replace command along with the mask target seem to be dropping the unreplaced values and I can't see how to remedy this.

推荐答案

基于正则表达式的str.replace

这个带有str.replace的正则表达式模式应该很好.

Regex-based with str.replace

This regex pattern with str.replace should do nicely.

s.str.replace(r'\.(?=.*?\.)', '')

0    1234.5
1     123.5
2    2345.6
3     678.9
dtype: object

这个想法是，只要要替换的字符更多，就继续替换.这是使用的正则表达式的细分.

The idea is that, as long as there are more characters to replace, keep replacing. Here's a breakdown of the regular expression used.

\.     # '.'
(?=    # positive lookahead
.*?    # match anything
\.     # look for '.'
)

---

与np.vectorize

一起玩

如果要使用count进行此操作，这并非不可能，但这是一个挑战.您可以使用np.vectorize使其更容易.首先，定义一个函数

---

Fun with np.vectorize

If you want to do this using count, it isn't impossible, but it is a challenge. You can make this easier with np.vectorize. First, define a function,

def foo(r, c):
    return r.replace('.', '', c)

矢量化它，

v = np.vectorize(foo)

现在，调用函数v，并传递s和要替换的计数.

Now, call the function v, passing s and the counts to replace.

pd.Series(v(s, s.str.count(r'\.') - 1))

0    1234.5
1     123.5
2    2345.6
3     678.9
dtype: object

请记住，这基本上是一个光荣的循环.

Keep in mind that this is basically a glorified loop.

与vectorize等效的python是

The python equivalent of vectorize would be,

r = []
for x, y in zip(s, s.str.count(r'\.') - 1):
    r.append(x.replace('.', '', y))

pd.Series(r)

0    1234.5
1     123.5
2    2345.6
3     678.9
dtype: object

或者，使用列表推导:

pd.Series([x.replace('.', '', y) for x, y in zip(s, s.str.count(r'\.') - 1)])

0    1234.5
1     123.5
2    2345.6
3     678.9
dtype: object

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