CODEWITHSUNDEEP
pythonpandas
arcee123
arcee123

Reputation: 243

import list as values of a pandas dataframe

If I have this list:

list = ['a', 'b', 'c', 'd']

and I convert to a dataframe using:

record = pd.DataFrame(list)

I get this:

     0
0    a
1    b
2    c
3    d

How do I get this instead, thanks??

     0    1    2    3
0    a    b    c    d

Upvotes: 0

Views: 265

Answers (4)

David Buck
David Buck

Reputation: 3830

You can transpose it into that format with T

record = pd.DataFrame(data).T

I liked the look of the other ([data]) solution posted, so I thought I'd check which was fastest for interest:

%timeit record = pd.DataFrame(data).T
820 µs ± 181 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit record = pd.DataFrame([data])
981 µs ± 323 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

So it appears that Pandas T does have a performance advantage over building the list within a list first. This held true with 400 entries in the list as well.

And as noted elsewhere - it's not good practice to redefine list as it's a system keyword.

Upvotes: 1

delete
delete

Reputation: 75

This should work:

import pandas as pd

L = ['a','b','c','d']

df = pd.DataFrame([L])
print(df)

Output:

  0 1 2 3
0 a b c d

I recommend you don't use the word "list" to name your list.

Upvotes: 1

CEWeinhauer
CEWeinhauer

Reputation: 123

You can just pass in list as [list]

list = ['a', 'b', 'c', 'd']
record = pd.DataFrame([list])

Upvotes: 2

Carles
Carles

Reputation: 2829

Use transpose() 

record = pd.DataFrame(list).transpose()
record
    Out[3]: 
       0  1  2  3
    0  a  b  c  d

Additionally, you can use: 

pd.DataFrame(list).T

Upvotes: 1

Related Questions