在pandas/python中的数据框中组合两列文本

Question

我在使用pandas的python中有一个20 x 4000的数据帧.其中两列名为Year和quarter.我想创建一个名为period的变量,将Year = 2000和quarter = q2变为2000q2

任何人都可以帮忙吗？

Answer 1

dataframe["period"] = dataframe["Year"].map(str) + dataframe["quarter"]

Answer 2

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

产生此数据帧

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

通过替换df[['Year', 'quarter']]数据帧的任何列切片,此方法可以概括为任意数量的字符串列,例如df.iloc[:,0:2].apply(lambda x: ''.join(x), axis=1).

您可以在此处查看有关apply()方法的更多信息

Answer 3

小数据集(<150rows)

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

或略慢但更紧凑:

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

更大的数据集(> 150)

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

更新:时序图Pandas 0.23.4

让我们在200K行DF上测试它:

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:使用Pandas 0.19.0的新时间

没有CPU/GPU优化的时序(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优化的时序:

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

由@ anton-vbr回答贡献

Answer 4

该方法cat()的的.str访问可以很好地表现这一点:

>>> 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() 甚至允许你添加一个分隔符,例如,假设你只有年份和期间的整数,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

连接多个列只是传递一个系列列表或一个包含除第一列之外的所有列的数据帧作为str.cat()在第一列(系列)上调用的参数:

>>> 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具有空值,则需要包含参数na_rep以使用字符串替换NaN值,否则组合列将默认为NaN.

Answer 5

这次使用lamba函数和string.format().

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

这允许您根据需要使用非字符串并重新格式化值.

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

Answer 6

虽然@silvado答案是好的,如果你改变df.map(str)到df.astype(str)它会更快:

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

In [131]: %timeit df["Year"].map(str)
10000 loops, best of 3: 132 us per loop

In [132]: %timeit df["Year"].astype(str)
10000 loops, best of 3: 82.2 us per loop

Answer 7

让我们假设你的 dataframe是df柱子Year和Quarter.

import pandas as pd
df = pd.DataFrame({'Quarter':'q1 q2 q3 q4'.split(), 'Year':'2000'})

假设我们想要查看数据帧;

df
>>>  Quarter    Year
   0    q1      2000
   1    q2      2000
   2    q3      2000
   3    q4      2000

最后,连接Year和Quarter如下.

df['Period'] = df['Year'] + ' ' + df['Quarter']

您现在print df 可以查看生成的数据框.

df
>>>  Quarter    Year    Period
    0   q1      2000    2000 q1
    1   q2      2000    2000 q2
    2   q3      2000    2000 q3
    3   q4      2000    2000 q4

如果您不想要年份和季度之间的空间,只需将其删除;

df['Period'] = df['Year'] + df['Quarter']

Answer 8

当您的数据插入到数据框中时,此命令应该可以解决您的问题:

df['period'] = df[['Year', 'quarter']].apply(lambda x: ' '.join(x.astype(str)), axis=1)

Answer 9

您可以使用 lambda：

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

然后将其用于创建新列：

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

Answer 10

这是一个我发现非常通用的实现:

In [1]: import pandas as pd 

In [2]: df = pd.DataFrame([[0, 'the', 'quick', 'brown'],
   ...:                    [1, 'fox', 'jumps', 'over'], 
   ...:                    [2, 'the', 'lazy', 'dog']],
   ...:                   columns=['c0', 'c1', 'c2', 'c3'])

In [3]: def str_join(df, sep, *cols):
   ...:     from functools import reduce
   ...:     return reduce(lambda x, y: x.astype(str).str.cat(y.astype(str), sep=sep), 
   ...:                   [df[col] for col in cols])
   ...: 

In [4]: df['cat'] = str_join(df, '-', 'c0', 'c1', 'c2', 'c3')

In [5]: df
Out[5]: 
   c0   c1     c2     c3                cat
0   0  the  quick  brown  0-the-quick-brown
1   1  fox  jumps   over   1-fox-jumps-over
2   2  the   lazy    dog     2-the-lazy-dog

Answer 11

效率更高

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

这是一个时间测试:

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()

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

Answer 12

此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个列。这不仅适用于字符串，而且适用于所有类型的列数据类型

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

Answer 13

归纳为多列，为什么不这样做：

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

Answer 14

您问题的简单答案。

    Year    quarter
0   2000    q1
1   2000    q2

> final_string = df['Year'] + '' + df['quarter']

> print(final_string)
  2000q1
  2000q2

Answer 15

使用zip甚至可以更快：

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

图形：

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()

Answer 16

这是我对上述解决方案的总结，这些解决方案使用列值之间的分隔符将具有 int 和 str 值的两列连接/组合到一个新列中。三种解决方案可用于此目的。

# 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)

Answer 17

我的看法....

listofcols = ['col1','col2','col3']
df['combined_cols'] = ''

for column in listofcols:
    df['combined_cols'] = df['combined_cols'] + ' ' + df[column]
'''