如何从熊猫的两列形成元组列

我有一个熊猫DataFrame，我想结合'lat'和'long'列形成一个元组。

<class 'pandas.core.frame.DataFrame'>
Int64Index: 205482 entries, 0 to 209018
Data columns:
Month           205482  non-null values
Reported by     205482  non-null values
Falls within    205482  non-null values
Easting         205482  non-null values
Northing        205482  non-null values
Location        205482  non-null values
Crime type      205482  non-null values
long            205482  non-null values
lat             205482  non-null values
dtypes: float64(4), object(5)

我尝试使用的代码是:

def merge_two_cols(series):
return (series['lat'], series['long'])


sample['lat_long'] = sample.apply(merge_two_cols, axis=1)

但是，这返回了以下错误:

---------------------------------------------------------------------------
AssertionError                            Traceback (most recent call last)
<ipython-input-261-e752e52a96e6> in <module>()
2     return (series['lat'], series['long'])
3
----> 4 sample['lat_long'] = sample.apply(merge_two_cols, axis=1)
5

.．.

AssertionError: Block shape incompatible with manager

我该如何解决这个问题呢?

183041 次浏览
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In [10]: df
Out[10]:
A         B       lat      long
0  1.428987  0.614405  0.484370 -0.628298
1 -0.485747  0.275096  0.497116  1.047605
2  0.822527  0.340689  2.120676 -2.436831
3  0.384719 -0.042070  1.426703 -0.634355
4 -0.937442  2.520756 -1.662615 -1.377490
5 -0.154816  0.617671 -0.090484 -0.191906
6 -0.705177 -1.086138 -0.629708  1.332853
7  0.637496 -0.643773 -0.492668 -0.777344
8  1.109497 -0.610165  0.260325  2.533383
9 -1.224584  0.117668  1.304369 -0.152561


In [11]: df['lat_long'] = df[['lat', 'long']].apply(tuple, axis=1)


In [12]: df
Out[12]:
A         B       lat      long                             lat_long
0  1.428987  0.614405  0.484370 -0.628298      (0.484370195967, -0.6282975278)
1 -0.485747  0.275096  0.497116  1.047605      (0.497115615839, 1.04760475074)
2  0.822527  0.340689  2.120676 -2.436831      (2.12067574274, -2.43683074367)
3  0.384719 -0.042070  1.426703 -0.634355      (1.42670326172, -0.63435462504)
4 -0.937442  2.520756 -1.662615 -1.377490     (-1.66261469102, -1.37749004179)
5 -0.154816  0.617671 -0.090484 -0.191906  (-0.0904840623396, -0.191905582481)
6 -0.705177 -1.086138 -0.629708  1.332853     (-0.629707821728, 1.33285348929)
7  0.637496 -0.643773 -0.492668 -0.777344   (-0.492667604075, -0.777344111021)
8  1.109497 -0.610165  0.260325  2.533383        (0.26032456699, 2.5333825651)
9 -1.224584  0.117668  1.304369 -0.152561     (1.30436900612, -0.152560909725)
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最佳答案

熟悉zip。它在处理列数据时非常方便。

df['new_col'] = list(zip(df.lat, df.long))

它比使用apply或map更简单、更快。像np.dstack这样的东西的速度是zip的两倍，但不会给你元组。

小开

Pandas有itertuples方法来做到这一点:

list(df[['lat', 'long']].itertuples(index=False, name=None))
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我想添加df.values.tolist()。(只要你不介意获得一列列表而不是元组)

import pandas as pd
import numpy as np


size = int(1e+07)
df = pd.DataFrame({'a': np.random.rand(size), 'b': np.random.rand(size)})


%timeit df.values.tolist()
1.47 s ± 38.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)


%timeit list(zip(df.a,df.b))
1.92 s ± 131 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
小开

你应该尝试使用pd.to_records(index=False):

import pandas as pd
df = pd.DataFrame({'language': ['en', 'ar', 'es'], 'greeting': ['Hi', 'اهلا', 'Hola']})
df


language  greeting
0       en    Hi
1       ar    اهلا
2       es   Hola


df['list_of_tuples'] = list(df[['language', 'greeting']].to_records(index=False))
df['list_of_tuples']


0    [en, Hi]
1    [ar, اهلا]
2    [es, Hola]

享受吧!


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