映射Serie和Dataframe值

对于循环

import pandas as pd
import numpy as np

df = pd.DataFrame(np.random.randint(0,10, (1000,3)), 
                  columns=['col_A', 'col_B', 'col_C'])

s = pd.Series([20,0,300], 
              index=['col_A', 'col_B', 'col_C'])

for col, idx in s.iteritems():
    df.loc[:idx, col] = -1

df
     col_A  col_B  col_C
0     -999   -999   -999
1     -999      3   -999
2     -999      3   -999
3     -999      2   -999
4     -999      1   -999
..     ...    ...    ...
995      2      6      9
996      1      9      5
997      2      6      4
998      4      0      1
999      9      2      8

熊猫适用

def f(c, s):
    c[:s[c.name]] = -1
    return c

df = df.apply(lambda c: f(c,s))
df 
     col_A  col_B  col_C
0       -1      6     -1
1       -1      1     -1
2       -1      6     -1
3       -1      1     -1
4       -1      6     -1
..     ...    ...    ...
995      2      4      3
996      2      0      0
997      8      5      7
998      3      5      5
999      5      7      7

性能在我的本地计算机上，使用N = 1000，for循环略快。将N增加到1M，apply方法更快：

def for_loop(N):
    df = pd.DataFrame(np.random.randint(0,10, (N,3)), 
                  columns=['col_A', 'col_B', 'col_C'])
    for col, idx in s.iteritems():
        df.loc[:idx, col] = -1
    return df

def apply_method(N):

    def f(c, s):
        c[:s[c.name]] = -1
        return c

    return df.apply(lambda c: f(c,s))

%timeit for_loop(1000)
1.19 ms ± 58.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit apply_method(1000)
185 ms ± 44.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit for_loop(1000_000)
303 ms ± 25.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit apply_method(1000_000)
162 ms ± 8.84 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

回答来源：stackoverflow