由于pandas的字符串方法没有优化（尽管pandas 2.0似乎不再是这样），如果你追求性能，最好在循环中使用Python字符串方法（这是在C中编译的）。似乎对每个字符串进行直接循环可能会给予最佳性能。

def evaluater(s):
    total, curr = 0, ''
    for e in s:
        # if a number concatenate to the previous number
        if e.isdigit():
            curr += e
        # if a string, look up its value in KeyDict
        # and multiply the currently collected number by it
        # and add to the total
        else:
            total += int(curr) * KeyDict[e]
            curr = ''
    return total

KeyDict = {'M': 1, 'N': 1, 'I': -1, 'S': 0, 'D': 1}
df['val'] = df['code'].map(evaluater)

性能：

KeyDict1 = {'M':'*1+','N':'*1+','I':'*-1+','S':'*0+','D':'*1+'}
df = pd.DataFrame({'code':['100M','60M10N40M','5S99M','1S25I100M','1D1S1I200M']*1000})

%timeit df.assign(val=df['code'].map(evaluater))
# 12.2 ms ± 579 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit df.assign(val=df['code'].apply(String2Val))    # @Marcelo Paco
# 61.8 ms ± 2.47 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit df.assign(val=df['code'].replace(KeyDict1, regex=True).str.rstrip('+').apply(pd.eval))   # @Ynjxsjmh
# 4.86 s ± 155 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

注意：你已经实现了类似的东西，但外部循环（for key, value in KeyDict.items()）是不必要的;因为KeyDict是一个字典，所以将其用作查找表;不要循环。另外，当只有一列相关时，.apply(axis=1)是一种非常糟糕的循环方式。选择该列并调用apply()。

您可以将加法符号添加到KeyDict的值，然后将code列的值替换为KeyDict，最后调用pd.eval进行计算。

KeyDict = {'M':'*1+','N':'*1+','I':'*-1+','S':'*0+','D':'*1+'}

df['val'] = (df['code'].replace(KeyDict, regex=True)
             .str.rstrip('+').apply(pd.eval))
# or you can use native python for loop since Series.apply is not efficient
df['val'] = [pd.eval(val) for val in df['code'].replace(KeyDict, regex=True).str.rstrip('+')]

print(df)

         code  val
0        100M  100
1   60M10N40M  110
2       5S99M   99
3   1S25I100M   75
4  1D1S1I200M  200

您可以尝试以下使用replace()的解决方案：

import pandas as pd

def String2Val(row):
    # Use replace to find an replace characters according to your KeyDict definition
    val = row.replace('M', '*1+').replace('N', '*1+').replace('I', '*-1+').replace('S', '*0+').replace('D', '*1+')
    # Ensure the last part of the string isn't a +
    if val[-1] == "+":
        # If it is, remove the + from the end
        val = val[:-1]
    # Return the evaluated value
    return eval(val)

df = pd.DataFrame({'code':['100M','60M10N40M','5S99M','1S25I100M','1D1S1I200M']})
# Modify it to use apply only on the code column. Which removes the need to use lambda and axis=1
df['Val'] = df['code'].apply(String2Val)

df：

code  Val
0        100M  100
1   60M10N40M  110
2       5S99M   99
3   1S25I100M   75
4  1D1S1I200M  200

另一种可能的解决方案是用相应的乘法因子替换字母，然后用eval计算字符串：

df['val'] = (df['code'].str.replace('M|N|D', '*1+', regex=True)
             .str.replace('I', '*(-1)+', regex=True)
             .str.replace('S', '*0+', regex=True)
             .str.replace(r'\+$', '', regex=True).map(eval))

输出：

code  val
0        100M  100
1   60M10N40M  110
2       5S99M   99
3   1S25I100M   75
4  1D1S1I200M  200

谢谢大家。如果这可能对其他人有帮助，我检查了每个人的解决方案在不同大小的数据上的性能，并获得了以下结果。非常有教育意义，在处理大型数据集时应用的伟大技巧。