类似于@geher的答案，但可以使用任何你喜欢的分隔符:

SEP = " "
INPUT_COLUMNS_WITH_SEP = ",sep,".join(INPUT_COLUMNS).split(",")

df.assign(sep=SEP)[INPUT_COLUMNS_WITH_SEP].sum(axis=1)

这次使用了string.format()的lamba函数。

import pandas as pd
df = pd.DataFrame({'Year': ['2014', '2015'], 'Quarter': ['q1', 'q2']})
print df
df['YearQuarter'] = df[['Year','Quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1)
print df

  Quarter  Year
0      q1  2014
1      q2  2015
  Quarter  Year YearQuarter
0      q1  2014      2014q1
1      q2  2015      2015q2

这允许您使用非字符串并根据需要重新格式化值。

import pandas as pd
df = pd.DataFrame({'Year': ['2014', '2015'], 'Quarter': [1, 2]})
print df.dtypes
print df

df['YearQuarter'] = df[['Year','Quarter']].apply(lambda x : '{}q{}'.format(x[0],x[1]), axis=1)
print df

Quarter     int64
Year       object
dtype: object
   Quarter  Year
0        1  2014
1        2  2015
   Quarter  Year YearQuarter
0        1  2014      2014q1
1        2  2015      2015q2

使用.combine_first。

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

下面是我对上述解决方案的总结，将两个具有int和str值的列连接/组合成一个新列，在列值之间使用分隔符。有三种解决方案可以达到这个目的。

# be cautious about the separator, some symbols may cause "SyntaxError: EOL while scanning string literal".
# e.g. ";;" as separator would raise the SyntaxError

separator = "&&" 

# pd.Series.str.cat() method does not work to concatenate / combine two columns with int value and str value. This would raise "AttributeError: Can only use .cat accessor with a 'category' dtype"

df["period"] = df["Year"].map(str) + separator + df["quarter"]
df["period"] = df[['Year','quarter']].apply(lambda x : '{} && {}'.format(x[0],x[1]), axis=1)
df["period"] = df.apply(lambda x: f'{x["Year"]} && {x["quarter"]}', axis=1)

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。

泛化到多个列，为什么不呢:

columns = ['whatever', 'columns', 'you', 'choose']
df['period'] = df[columns].astype(str).sum(axis=1)