How to create new columns in pandas dataframe with ONLY a list-comprehension?

Question

I want to create a set of n columns in a DataFrame each assigned a separate value using a list comprehension.

#My original dataframe
df = pd.DataFrame({'A':[1,2,3],'B':[4,5,6]})

   A  B
0  1  4
1  2  5
2  3  6

#Expected output - 

pd.concat([df, pd.DataFrame(np.tile(np.array([5,10,15,20,25])[:,None], 3).T)], axis=1)

   A  B  0   1   2   3   4
0  1  4  5  10  15  20  25
1  2  5  5  10  15  20  25
2  3  6  5  10  15  20  25

I need to do it in this fashion -

#ROUGH structure of the code that I am looking for -
n = "number of columns i want to add"
df[[i for i in range(n)]] = numpyarray #whose shape is (n,3)

The error that I face is quite obvious -

KeyError: "None of [Int64Index([0, 1, 2], dtype='int64')] are in the [columns]"

#AND

SyntaxError: can't assign to list comprehension

---

I have read other solutions which allow adding multiple columns but this one specifically needs a loop with an iterator of n because -

1. The data frame may need 25 columns added and that doesn't depend on the array of values
2. The array of values can be (3, 15) which means that last 10 of the columns will not take their values from the array
3. The prefered solution would be a list comprehension since the list of columns that I would be creating (25 for example) come from a list comprehension based iterator

Answer 1

One idea for create columns by list comprehension, tested in pandas 1.1.1:

df = pd.DataFrame({'A':[1,2,3],'B':[4,5,6]})

#list created by list comprehension
L = [i + 1 for i in range(5)]
print (L)
[1, 2, 3, 4, 5]

n = len(L)
df[list(range(n))] = L

print (df)
   A  B  0  1  2  3  4
0  1  4  1  2  3  4  5
1  2  5  1  2  3  4  5
2  3  6  1  2  3  4  5

Answer 2

Here's an updated version of the solution.

import pandas as pd
import numpy as np
df = pd.DataFrame({'A':[1,2,3],'B':[4,5,6]})
print(df)
n = 10

df = pd.concat([df,pd.DataFrame(
    np.tile([5*(i+1) for i in range(n)],len(df)).reshape(len(df),n),
    columns=[i+1 for i in range (n)])],axis=1)

print(df)

The output from this is as follows:

Original DataFrame:

   A  B
0  1  4
1  2  5
2  3  6

Merged dataframe

   A  B  1   2   3   4   5   6   7   8   9  10
0  1  4  5  10  15  20  25  30  35  40  45  50
1  2  5  5  10  15  20  25  30  35  40  45  50
2  3  6  5  10  15  20  25  30  35  40  45  50

We need to get a table with values [5,10,15,...,n*5]. To achieve this, I am using:

np.tile([5*(i+1) for i in range(n)],len(df))

This will give me an array like this:

array([ 5, 10, 15, 20, 25, 30, 35, 40, 45, 50,  5, 10, 15, 20, 25, 30, 35,
       40, 45, 50,  5, 10, 15, 20, 25, 30, 35, 40, 45, 50])

Now we need to switch this to 3 rows by n columns where n=10 in this example. I am doing that using:

reshape(len(df),n)

Here len(df) = 3 and n = 10

The result of

np.tile([5*(i+1) for i in range(n)],len(df)).reshape(len(df),n)

will be :

array([[ 5, 10, 15, 20, 25, 30, 35, 40, 45, 50],
       [ 5, 10, 15, 20, 25, 30, 35, 40, 45, 50],
       [ 5, 10, 15, 20, 25, 30, 35, 40, 45, 50]])

Now that I have the values listed, I just need to get the column names. I am using a list comprehension to create the column names.

columns=[i+1 for i in range (n)])]

And obviously we got to use axis=1 otherwise it will not concatenate correctly.

Putting all this together gives you the final result set.

I went back and tried to use Akshay's logic. Here' what I got. This also works.

df2 = pd.concat([df,pd.DataFrame(
    np.tile(np.array([[5*i] for i in range(1,n+1)]), len(df)).T,
    columns=[i+1 for i in range (n)])],axis=1)
print(df2)

If you think there are easier ways to do this, please let me know so I can learn as well.

The previous response is below:

I am fairly new to pandas and still learning to figure things out. Here's what I tried and it looks like this is what you want.

import pandas as pd
df = pd.DataFrame({'A':[1,2,3],'B':[4,5,6]})
lst = [5,10,15,20,25]
n = 6
for i in range(1,n): df[i] =lst[i-1]
print(df)

This gave me the following output:

   A  B  1   2   3   4   5
0  1  4  5  10  15  20  25
1  2  5  5  10  15  20  25
2  3  6  5  10  15  20  25

Does this make sense and is this what you are looking for?