你可以使用lambda:

combine_lambda = lambda x: '{}{}'.format(x.Year, x.quarter)

然后使用它来创建新列:

df['period'] = df.apply(combine_lambda, axis = 1)

如果两个列都是字符串，你可以直接连接它们:

df["period"] = df["Year"] + df["quarter"]

如果一个(或两个)列不是字符串类型的，你应该先转换它(它们)，

df["period"] = df["Year"].astype(str) + df["quarter"]

这样做时要当心nan !

如果你需要连接多个字符串列，你可以使用agg:

df['period'] = df[['Year', 'quarter', ...]].agg('-'.join, axis=1)

其中“-”是分隔符。

使用zip可以更快:

df["period"] = [''.join(i) for i in zip(df["Year"].map(str),df["quarter"])]

图:

import pandas as pd
import numpy as np
import timeit
import matplotlib.pyplot as plt
from collections import defaultdict

df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']})

myfuncs = {
"df['Year'].astype(str) + df['quarter']":
    lambda: df['Year'].astype(str) + df['quarter'],
"df['Year'].map(str) + df['quarter']":
    lambda: df['Year'].map(str) + df['quarter'],
"df.Year.str.cat(df.quarter)":
    lambda: df.Year.str.cat(df.quarter),
"df.loc[:, ['Year','quarter']].astype(str).sum(axis=1)":
    lambda: df.loc[:, ['Year','quarter']].astype(str).sum(axis=1),
"df[['Year','quarter']].astype(str).sum(axis=1)":
    lambda: df[['Year','quarter']].astype(str).sum(axis=1),
    "df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1)":
    lambda: df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1),
    "[''.join(i) for i in zip(dataframe['Year'].map(str),dataframe['quarter'])]":
    lambda: [''.join(i) for i in zip(df["Year"].map(str),df["quarter"])]
}

d = defaultdict(dict)
step = 10
cont = True
while cont:
    lendf = len(df); print(lendf)
    for k,v in myfuncs.items():
        iters = 1
        t = 0
        while t < 0.2:
            ts = timeit.repeat(v, number=iters, repeat=3)
            t = min(ts)
            iters *= 10
        d[k][lendf] = t/iters
        if t > 2: cont = False
    df = pd.concat([df]*step)

pd.DataFrame(d).plot().legend(loc='upper center', bbox_to_anchor=(0.5, -0.15))
plt.yscale('log'); plt.xscale('log'); plt.ylabel('seconds'); plt.xlabel('df rows')
plt.show()

正如前面提到的，必须将每个列转换为字符串，然后使用加号运算符将两个字符串列合并。使用NumPy可以大大提高性能。

%timeit df['Year'].values.astype(str) + df.quarter
71.1 ms ± 3.76 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit df['Year'].astype(str) + df['quarter']
565 ms ± 22.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']})
df['period'] = df[['Year', 'quarter']].apply(lambda x: ''.join(x), axis=1)

产生这个数据框架

   Year quarter  period
0  2014      q1  2014q1
1  2015      q2  2015q2

该方法通过将df[['Year'， 'quarter']]替换为数据帧的任意列片来泛化任意数量的字符串列，例如df.iloc[:，0:2]。Apply (lambda x: " .join(x)， axis=1)。

您可以在这里查看有关apply()方法的更多信息

小数据集(< 150行)

[''.join(i) for i in zip(df["Year"].map(str),df["quarter"])]

或者稍慢但更紧凑:

df.Year.str.cat(df.quarter)

更大的数据集(> 150rows)

df['Year'].astype(str) + df['quarter']

更新:定时图熊猫0.23.4

让我们在200K行上测试一下:

In [250]: df
Out[250]:
   Year quarter
0  2014      q1
1  2015      q2

In [251]: df = pd.concat([df] * 10**5)

In [252]: df.shape
Out[252]: (200000, 2)

更新:新的计时使用熊猫0.19.0

没有CPU/GPU优化的计时(从最快到最慢排序):

In [107]: %timeit df['Year'].astype(str) + df['quarter']
10 loops, best of 3: 131 ms per loop

In [106]: %timeit df['Year'].map(str) + df['quarter']
10 loops, best of 3: 161 ms per loop

In [108]: %timeit df.Year.str.cat(df.quarter)
10 loops, best of 3: 189 ms per loop

In [109]: %timeit df.loc[:, ['Year','quarter']].astype(str).sum(axis=1)
1 loop, best of 3: 567 ms per loop

In [110]: %timeit df[['Year','quarter']].astype(str).sum(axis=1)
1 loop, best of 3: 584 ms per loop

In [111]: %timeit df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1)
1 loop, best of 3: 24.7 s per loop

使用CPU/GPU优化计时:

In [113]: %timeit df['Year'].astype(str) + df['quarter']
10 loops, best of 3: 53.3 ms per loop

In [114]: %timeit df['Year'].map(str) + df['quarter']
10 loops, best of 3: 65.5 ms per loop

In [115]: %timeit df.Year.str.cat(df.quarter)
10 loops, best of 3: 79.9 ms per loop

In [116]: %timeit df.loc[:, ['Year','quarter']].astype(str).sum(axis=1)
1 loop, best of 3: 230 ms per loop

In [117]: %timeit df[['Year','quarter']].astype(str).sum(axis=1)
1 loop, best of 3: 230 ms per loop

In [118]: %timeit df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1)
1 loop, best of 3: 9.38 s per loop

回答@anton-vbr的贡献