Take multiple lists into dataframe

How do I take multiple lists and put them as different columns in a python dataframe? I tried this solution but had some trouble.

Attempt 1:

• Have three lists, and zip them together and use that res = zip(lst1,lst2,lst3)
• Yields just one column

Attempt 2:

percentile_list = pd.DataFrame({'lst1Tite' : [lst1],
                                'lst2Tite' : [lst2],
                                'lst3Tite' : [lst3] }, 
                                columns=['lst1Tite','lst1Tite', 'lst1Tite'])

• yields either one row by 3 columns (the way above) or if I transpose it is 3 rows and 1 column

How do I get a 100 row (length of each independent list) by 3 column (three lists) pandas dataframe?

Answers:

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Method 1

I think you’re almost there, try removing the extra square brackets around the lst‘s (Also you don’t need to specify the column names when you’re creating a dataframe from a dict like this):

import pandas as pd
lst1 = range(100)
lst2 = range(100)
lst3 = range(100)
percentile_list = pd.DataFrame(
    {'lst1Title': lst1,
     'lst2Title': lst2,
     'lst3Title': lst3
    })

percentile_list
    lst1Title  lst2Title  lst3Title
0          0         0         0
1          1         1         1
2          2         2         2
3          3         3         3
4          4         4         4
5          5         5         5
6          6         6         6
...

If you need a more performant solution you can use np.column_stack rather than zip as in your first attempt, this has around a 2x speedup on the example here, however comes at bit of a cost of readability in my opinion:

import numpy as np
percentile_list = pd.DataFrame(np.column_stack([lst1, lst2, lst3]), 
                               columns=['lst1Title', 'lst2Title', 'lst3Title'])

Method 2

Adding to Aditya Guru‘s answer here. There is no need of using map. You can do it simply by:

pd.DataFrame(list(zip(lst1, lst2, lst3)))

This will set the column’s names as 0,1,2. To set your own column names, you can pass the keyword argument columns to the method above.

pd.DataFrame(list(zip(lst1, lst2, lst3)),
              columns=['lst1_title','lst2_title', 'lst3_title'])

Method 3

Adding one more scalable solution.

lists = [lst1, lst2, lst3, lst4]
df = pd.concat([pd.Series(x) for x in lists], axis=1)

Method 4

Just adding that using the first approach it can be done as –

pd.DataFrame(list(map(list, zip(lst1,lst2,lst3))))

Method 5

There are several ways to create a dataframe from multiple lists.

list1=[1,2,3,4]
list2=[5,6,7,8]
list3=[9,10,11,12]

1. pd.DataFrame({'list1':list1, 'list2':list2, 'list3'=list3})
2. pd.DataFrame(data=zip(list1,list2,list3),columns=['list1','list2','list3'])

Method 6

Adding to above answers, we can create on the fly

df= pd.DataFrame()
list1 = list(range(10))
list2 = list(range(10,20))
df['list1'] = list1
df['list2'] = list2
print(df)

hope it helps !

Method 7

@oopsi used pd.concat() but didn’t include the column names. You could do the following, which, unlike the first solution in the accepted answer, gives you control over the column order (avoids dicts, which are unordered):

import pandas as pd
lst1 = range(100)
lst2 = range(100)
lst3 = range(100)

s1=pd.Series(lst1,name='lst1Title')
s2=pd.Series(lst2,name='lst2Title')
s3=pd.Series(lst3 ,name='lst3Title')
percentile_list = pd.concat([s1,s2,s3], axis=1)

percentile_list
Out[2]: 
    lst1Title  lst2Title  lst3Title
0           0          0          0
1           1          1          1
2           2          2          2
3           3          3          3
4           4          4          4
5           5          5          5
6           6          6          6
7           7          7          7
8           8          8          8
...

Method 8

you can simple use this following code

train_data['labels']= train_data[["LABEL1","LABEL1","LABEL2","LABEL3","LABEL4","LABEL5","LABEL6","LABEL7"]].values.tolist()
train_df = pd.DataFrame(train_data, columns=['text','labels'])

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0