使用扩展正则表达式替换：

import re

pat = re.compile(r'^(\d+\.)([\d.]+)')  # precompiled pattern
repl = lambda m: m.group(1) + m.group(2).replace(".", "")
df.A = pd.to_numeric(df.A.str.replace(pat, repl))
df.B = pd.to_numeric(df.B.str.replace(pat, repl))

A          B
0   54.646400  46.848400
1   64.684610  65.400000
2   95.795270  65.644000
3  484.644161  45.450000
4   71.257900  21.100000
5   12.800000  10.800000
6    9.600000  12.500000
7  312.400000  12.787577

与Leonid的答案非常相似，但没有使用. apply。不确定哪个是最好的。

import pandas as pd

data = {
    'A': ['54.6464', '64.68461', '95.79527', '484.644.161', '71.257.9', '12.8', '9.6', '312.4'],
    'B': ['46.8484', '65.4', '65.644', '45.45', '21.1', '10.8', '12.5', '12.787.57.674']
}

df = pd.DataFrame(data=data)

for key in df:
    df[key] = [x.split('.')[0]+'.'+''.join(x.split('.')[1:]) for x in df[key].tolist()]
    
print(df)
            A            B
0     54.6464      46.8484
1    64.68461         65.4
2    95.79527       65.644
3  484.644161        45.45
4     71.2579         21.1
5        12.8         10.8
6         9.6         12.5
7       312.4  12.78757674

这是使用正则表达式删除第二个小数点的解决方案，而不是@Leonid Astrain提出的函数，看起来像缩短版：

import pandas as pd

# create a sample dataframe
data = {
    'A': ['54.6464', '64.68461', '95.79527', '484.644.161', '71.257.9', '12.8', '9.6', '312.4'],
    'B': ['46.8484', '65.4', '65.644', '45.45', '21.1', '10.8', '12.5', '12.787.57.674']
}
df = pd.DataFrame(data)

# use regex to remove second decimal point
df = df.replace(r'\.(?=.*\.)', '', regex=True)

# convert columns to float
df['A'] = df['A'].astype(float)
df['B'] = df['B'].astype(float)

print(df)

输出将为：

A          B
0   54.646400  46.848400
1   64.684610  65.400000
2   95.795270  65.644000
3  484.644161  45.450000
4   71.257900  21.100000
5   12.800000  10.800000
6    9.600000  12.500000
7  312.400000  12.787577

我会这样做：

def rectify_decimal(string):
   parts = string.split('.')
   if len(parts) > 1:
     return f"{parts[0]}.{''.join(parts[1:])}"
   else:
     return(parts[0])

df['A'] = df['A'].apply(rectify_decimal)

如果始终存在小数点：

df['A'].str.split('.').str[0] + '.' + df['A'].str.split('.').str[1:].str.join('')

您可以使用一些字符串操作：

def convert(sr):
    return (sr.str.split('.', n=1, expand=True)
              .pipe(lambda x: x[0] + '.' + x[1].str.replace('.', '', regex=False))
              .astype(float))

df = df.apply(convert)
print(df)

# Output
            A          B
0   54.646400  46.848400
1   64.684610  65.400000
2   95.795270  65.644000
3  484.644161  45.450000
4   71.257900  21.100000
5   12.800000  10.800000
6    9.600000  12.500000
7  312.400000  12.787577