通过zip从两个列表创建的字典使用

map,最后在

fillna之前替换为原始的非匹配值：

d = dict(zip(L2, L5))

print (d)

{-1: 9, -3: 10, -4: 11}

df['a'] = df['a'].map(d).fillna(df['a'])

print (df)

a b

0 9.0 23

1 10.0 45

2 11.0 67

3 11.0 89

4 10.0 0

5 4.0 -1

6 5.0 2

7 6.0 3

性能：

它取决于替换列表长度的值的数量：

列表长度为100：

np.random.seed(123)

N = 1000000

df = pd.DataFrame({'a':np.random.randint(1000, size=N)})

L2 = np.arange(100)

L5 = np.arange(100) + 10

In [336]: %timeit df['d'] = np.select([df['a'] == i for i in L2], L5, df['a'])

180 ms ± 1.07 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [337]: %timeit df['a'].map(dict(zip(L2, L5))).fillna(df['a'])

56.9 ms ± 2.55 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

如果列表长度很小(例如3)：

np.random.seed(123)

N = 1000000

df = pd.DataFrame({'a':np.random.randint(100, size=N)})

L2 = np.arange(3)

L5 = np.arange(3) + 10

In [339]: %timeit df['d'] = np.select([df['a'] == i for i in L2], L5, df['a'])

11.9 ms ± 40.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [340]: %timeit df['a'].map(dict(zip(L2, L5))).fillna(df['a'])

54 ms ± 215 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)