Question

假设我有以下返回元组的函数：

df = pd.DataFrame({'col1': [1,2,3]})
df['test'] = df['col1'].apply(return_tuple)

>>
    col1    test
0   1   (1, one)
1   2   (2, two)
2   one (1, one)

如果我使用apply方法，它将返回：

df['test_1'] = df['col1'].apply(return_tuple)??? # get 0-index in tuple
df['test_2'] = df['col1'].apply(return_tuple)??? # get 1 index in tuple

>>
    col1    test_1    test_2
0   1         1        one
1   2         2        two
2   one       1        one

但是我想要这样的东西：

/etc/wgetrc

谢谢。

Answer 1

使用unique_ptr和元组拆包，在亚历山大和拉兹迪的答案之间的某个地方：

<div class="section active" id="section0">
      <gwd-genericad id="gwd-ad">
        <gwd-pagedeck class="gwd-page-container" id="pagedeck">
          <gwd-page id="page1" class="gwd-page-wrapper gwd-page-size gwd-lightbox" data-gwd-width="100%" data-gwd-height="100%">
            <div class="gwd-page-content gwd-page-size">
              <gwd-image id="BG-01" source="assets/BG-01.png" scaling="stretch" class="gwd-image-1fyz" data-gwd-tl-locked=""></gwd-image>
              <gwd-image id="BG-01-01" source="assets/BG-01-01.png" scaling="stretch" class="gwd-image-q9o2" data-gwd-tl-locked=""></gwd-image>
              <gwd-image id="BG-01-01-01" source="assets/BG-01-01-01.png" scaling="stretch" class="gwd-image-qajr gwd-gen-phfngwdanimation"></gwd-image>
              <gwd-image id="BG-01-01-01-01" source="assets/BG-01-01-01-01.png" scaling="stretch" class="gwd-image-1d50 gwd-gen-19lhgwdanimation"></gwd-image><span class="gwd-span-1rx3 gwd-gen-148jgwdanimation">SAVE THE DATE</span><span class="gwd-span-wmst gwd-gen-qsz5gwdanimation">WILLIAM &amp; TIEN LI</span>
              <gwd-image id="arrow-down-sign-to-navigate-01" source="assets/arrow-down-sign-to-navigate-01.png" scaling="stretch" class="gwd-image-jolg gwd-gen-1kijgwdanimation"></gwd-image>
               </div>
            </gwd-page>
        </gwd-pagedeck>
      </gwd-genericad>
    </div>
    <div class="section" id="section1">
      <gwd-page id="page1_1" class="gwd-page-wrapper gwd-lightbox gwd-page-1mqd" data-gwd-width="100%" data-gwd-height="100%">
        <div class="gwd-page-content gwd-div-a9om">
          <gwd-image id="quaver" source="assets/quaver.png" scaling="stretch" class="gwd-image-ei8t"></gwd-image>
          <gwd-audio id="gwd-audio_1" sources="http://k007.kiwi6.com/hotlink/auu01qfyna/beautiful-in-white.mp3" class="gwd-audio-1q64" autoplay="" loop=""></gwd-audio>
        </div>
      </gwd-page>
    </div>

Answer 2

您也可以一步一步完成

import pandas as pd
df = pd.DataFrame({'col1': [1,2,3]})

def return_tuple(x):
    if x['col1'] in [1,'1','one']: 
        return pd.Series([1, 'one'])
    else:
        return pd.Series([2, 'two'])

df[['test_1', 'test_2']] = df.apply(return_tuple, axis=1)

Answer 3

import pandas as pd

def return_tuple(x):
    if x in [1, '1', 'one']:
        return 1, 'one'
    else:
        return 2, 'two'


df_1 = pd.DataFrame({'col1': [1, 2, 3]})

df_1['test_1'] = df_1['col1'].apply(lambda item: return_tuple(item)[0])
df_1['test_2'] = df_1['col1'].apply(lambda item: return_tuple(item)[1])

print(df_1)

就这么简单！

有关lambda函数的更多信息，请参见https://realpython.com/python-lambda/。有关SO的一些相关问题，例如this one。

Answer 4

您还可以在不更改现有功能的情况下完成此一线工作：

df[['test_1','test_2']] = pd.DataFrame(df['col1'].apply(return_tuple).tolist(),index=df.index)

Answer 5

经过一些修改，结果如下：

import pandas as pd

def return_tuple(x):
    if x in [1, '1', 'one']:
        return 1, 'one'
    else:
        return 2, 'two'

df_1 = pd.DataFrame({'col1': [1, 2, 3]})

df_1['test_1'], df_1['test_2'] = zip(*df_1['col1'].apply(return_tuple))

使用df连接

Answer 6

您可以将其扩展以执行所需的操作：

def return_tuple(x):
    if x in [1,'1','one']: 
        return (1, 'one')
    else:
        return (2, 'two')

df = pd.DataFrame({'col1': [1,2,3]})
df['test'] = df['col1'].apply(return_tuple)


 df[['test','test2']] = pd.DataFrame(df['test'].to_list(), index=df.index)


Out[32]: 
   col1  test test2
0     1     1   one
1     2     2   two
2     3     2   two

在熊猫中使用“应用”方法时，如何在元组中分配元素？

6 个答案: