I have a dataframe with one column and I would like to get a Dataframe with N columns all of which will be identical to the first one. I can simply duplicate it by:
df[['new column name']] = df[['column name']]
but I have to make more than 1000 identical columns that's why it doesnt work . One important thing is figures in columns should change for instance if first columns is 0 the nth column is n and the previous is n-1
If it's a single column, you can use tranpose and then simply replicate them with pd.concat and tranpose back to the original format, this avoids looping and should be faster, then you can change the column names in a second line, but without dealing with all the data in the dataframe which would be the most consuming performance wise:
import pandas as pd
df = pd.DataFrame({'Column':[1,2,3,4,5]})
Original dataframe:
Column
0 1
1 2
2 3
3 4
4 5
df = pd.concat([df.T]*1000).T
Output:
Column Column Column Column ... Column Column Column Column
0 1 1 1 1 ... 1 1 1 1
1 2 2 2 2 ... 2 2 2 2
2 3 3 3 3 ... 3 3 3 3
3 4 4 4 4 ... 4 4 4 4
4 5 5 5 5 ... 5 5 5 5
[5 rows x 1000 columns]
df.columns = ['Column'+'_'+str(i) for i in range(1000)]
Say that you have a df:, with column name 'company_name' that consists of 8 companies:
df = {"company_name":{"0":"Telia","1":"Proximus","2":"Tmobile","3":"Orange","4":"Telefonica","5":"Verizon","6":"AT&T","7":"Koninklijke"}}
company_name
0 Telia
1 Proximus
2 Tmobile
3 Orange
4 Telefonica
5 Verizon
6 AT&T
7 Koninklijke
You can use a loop and range to determine how many identical columns to create, and do:
for i in range(0,1000):
df['company_name'+str(i)] = df['company_name']
which results in the shape of the df:
df.shape
(8, 1001)
i.e. it replicated 1000 times the same columns. The names of the duplicated columns will be the same as the original one, plus an integer (=+1) at the end:
'company_name', 'company_name0', 'company_name1', 'company_name2','company_name..N'
df
A B C
0 x x x
1 y x z
Duplicate column "C" 5 times using df.assign:
n = 5
df2 = df.assign(**{f'C{i}': df['C'] for i in range(1, n+1)})
df2
A B C C1 C2 C3 C4 C5
0 x x x x x x x x
1 y x z z z z z z
Set n to 1000 to get your desired output.
You can also directly assign the result back:
df[[f'C{i}' for i in range(1, n+1)]] = df[['C']*n].to_numpy()
df
A B C C1 C2 C3 C4 C5
0 x x x x x x x x
1 y x z z z z z z
I think the most efficient is to index with DataFrame.loc instead of using an outer loop
n = 3
new_df = df.loc[:, ['column_duplicate']*n +
df.columns.difference(['column_duplicate']).tolist()]
print(new_df)
column_duplicate column_duplicate column_duplicate other
0 0 0 0 10
1 1 1 1 11
2 2 2 2 12
3 3 3 3 13
4 4 4 4 14
5 5 5 5 15
6 6 6 6 16
7 7 7 7 17
8 8 8 8 18
9 9 9 9 19
If you want add a suffix
suffix_tup = ('a', 'b', 'c')
not_dup_cols = df.columns.difference(['column_duplicate']).tolist()
new_df = (df.loc[:, ['column_duplicate']*len(suffix_tup) +
not_dup_cols]
.set_axis(list(map(lambda suffix: f'column_duplicate_{suffix}',
suffix_tup)) +
not_dup_cols, axis=1)
)
print(new_df)
column_duplicate_a column_duplicate_b column_duplicate_c other
0 0 0 0 10
1 1 1 1 11
2 2 2 2 12
3 3 3 3 13
4 4 4 4 14
5 5 5 5 15
6 6 6 6 16
7 7 7 7 17
8 8 8 8 18
or add an index
n = 3
not_dup_cols = df.columns.difference(['column_duplicate']).tolist()
new_df = (df.loc[:, ['column_duplicate']*n +
not_dup_cols]
.set_axis(list(map(lambda x: f'column_duplicate_{x}', range(n))) +
not_dup_cols, axis=1)
)
print(new_df)
column_duplicate_0 column_duplicate_1 column_duplicate_2 other
0 0 0 0 10
1 1 1 1 11
2 2 2 2 12
3 3 3 3 13
4 4 4 4 14
5 5 5 5 15
6 6 6 6 16
7 7 7 7 17
8 8 8 8 18
9 9 9 9 19