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)

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

.str访问器的cat()方法非常适用于此:

>>> import pandas as pd
>>> df = pd.DataFrame([["2014", "q1"], 
...                    ["2015", "q3"]],
...                   columns=('Year', 'Quarter'))
>>> print(df)
   Year Quarter
0  2014      q1
1  2015      q3
>>> df['Period'] = df.Year.str.cat(df.Quarter)
>>> print(df)
   Year Quarter  Period
0  2014      q1  2014q1
1  2015      q3  2015q3

Cat()甚至允许你添加分隔符，例如，假设你只有整数年和周期，你可以这样做:

>>> import pandas as pd
>>> df = pd.DataFrame([[2014, 1],
...                    [2015, 3]],
...                   columns=('Year', 'Quarter'))
>>> print(df)
   Year Quarter
0  2014       1
1  2015       3
>>> df['Period'] = df.Year.astype(str).str.cat(df.Quarter.astype(str), sep='q')
>>> print(df)
   Year Quarter  Period
0  2014       1  2014q1
1  2015       3  2015q3

连接多个列只是将一个序列列表或一个包含除第一列外的所有数据帧作为参数传递给在第一列(series)上调用的str.cat():

>>> df = pd.DataFrame(
...     [['USA', 'Nevada', 'Las Vegas'],
...      ['Brazil', 'Pernambuco', 'Recife']],
...     columns=['Country', 'State', 'City'],
... )
>>> df['AllTogether'] = df['Country'].str.cat(df[['State', 'City']], sep=' - ')
>>> print(df)
  Country       State       City                   AllTogether
0     USA      Nevada  Las Vegas      USA - Nevada - Las Vegas
1  Brazil  Pernambuco     Recife  Brazil - Pernambuco - Recife

请注意，如果您的pandas dataframe/series有空值，您需要包括参数na_rep来用字符串替换NaN值，否则合并的列将默认为NaN。

使用.combine_first。

df['Period'] = df['Year'].combine_first(df['Quarter'])

该解决方案使用中间步骤，将DataFrame的两列压缩为包含值列表的单列。 这不仅适用于字符串，而且适用于所有类型的列-dtype

import pandas as pd
df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']})
df['list']=df[['Year','quarter']].values.tolist()
df['period']=df['list'].apply(''.join)
print(df)

结果:

   Year quarter        list  period
0  2014      q1  [2014, q1]  2014q1
1  2015      q2  [2015, q2]  2015q2

更有效率的是

def concat_df_str1(df):
    """ run time: 1.3416s """
    return pd.Series([''.join(row.astype(str)) for row in df.values], index=df.index)

下面是一个时间测试:

import numpy as np
import pandas as pd

from time import time


def concat_df_str1(df):
    """ run time: 1.3416s """
    return pd.Series([''.join(row.astype(str)) for row in df.values], index=df.index)


def concat_df_str2(df):
    """ run time: 5.2758s """
    return df.astype(str).sum(axis=1)


def concat_df_str3(df):
    """ run time: 5.0076s """
    df = df.astype(str)
    return df[0] + df[1] + df[2] + df[3] + df[4] + \
           df[5] + df[6] + df[7] + df[8] + df[9]


def concat_df_str4(df):
    """ run time: 7.8624s """
    return df.astype(str).apply(lambda x: ''.join(x), axis=1)


def main():
    df = pd.DataFrame(np.zeros(1000000).reshape(100000, 10))
    df = df.astype(int)

    time1 = time()
    df_en = concat_df_str4(df)
    print('run time: %.4fs' % (time() - time1))
    print(df_en.head(10))


if __name__ == '__main__':
    main()

最后，当使用sum(concat_df_str2)时，结果不是简单的concat，它将转换为整数。