1

What would be the most efficient way to create a list of the labels from the dataframe below in the order of mylist?

import numpy as np
import pandas as pd    

mylist = ['a1.jpeg','a2.jpeg','b1.jpeg','b2.jpeg','c1.jpeg','c2.jpeg']
    
dat = np.array([(1, 2, 1, 1, 2, 2), ('a2jpeg', 'a1jpeg', 'c2jpeg', 'b2jpeg', 'b1jpeg' , 'c1jpeg')])
df = pd.DataFrame(np.transpose(dat), columns=['labels', 'filenames'])
df
>>labels    filenames
0   1   a2.jpeg
1   2   a1.jpeg
2   1   c2.jpeg
3   1   b2.jpeg
4   2   b1.jpeg
5   2   c1.jpeg
Angus Campbell
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2 Answers2

2

Just use sort_values:

>>> df.sort_values('filenames')
  labels filenames
1      2    a1jpeg
0      1    a2jpeg
4      2    b1jpeg
3      1    b2jpeg
5      2    c1jpeg
2      1    c2jpeg
>>>

To convert to list:

>>> df['filenames'].sort_values().tolist()
['a1jpeg', 'a2jpeg', 'b1jpeg', 'b2jpeg', 'c1jpeg', 'c2jpeg']
>>>
jezrael
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U13-Forward
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  • In what situation would it be more efficient to have a list and in what situations would it be better to output a tuple? – Angus Campbell Oct 11 '21 at 06:37
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    @AngusCampbell Read [this](https://stackoverflow.com/questions/626759/whats-the-difference-between-lists-and-tuples) for more info about that. – U13-Forward Oct 11 '21 at 06:39
2

Use Series.replace first and then change order by DataFrame.set_index with Series.reindex:

L = (df.assign(filenames = df['filenames'].replace('jpeg','.jpeg', regex=True))
       .set_index('filenames')['labels']
       .reindex(mylist)
       .tolist())
print (L)
['2', '1', '2', '1', '2', '1']

Or:

df['filenames'] = pd.Categorical(df['filenames'].replace('jpeg','.jpeg', regex=True), 
                                 ordered=True, 
                                 categories=mylist)

L = df.sort_values(by='filenames')['labels'].tolist()
print (L)
['2', '1', '2', '1', '2', '1']

If possible simplify solution by sorting in DataFrame.sort_values:

L = df.sort_values(by='filenames')['labels'].tolist()
print (L)
['2', '1', '2', '1', '2', '1']
jezrael
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