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)

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。

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)

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)

def madd(x):
    """Performs element-wise string concatenation with multiple input arrays.

    Args:
        x: iterable of np.array.

    Returns: np.array.
    """
    for i, arr in enumerate(x):
        if type(arr.item(0)) is not str:
            x[i] = x[i].astype(str)
    return reduce(np.core.defchararray.add, x)

例如:

data = list(zip([2000]*4, ['q1', 'q2', 'q3', 'q4']))
df = pd.DataFrame(data=data, columns=['Year', 'quarter'])
df['period'] = madd([df[col].values for col in ['Year', 'quarter']])

df

    Year    quarter period
0   2000    q1  2000q1
1   2000    q2  2000q2
2   2000    q3  2000q3
3   2000    q4  2000q4