Pandas DataFrame multiplication by array

Question

DataFrame

A B C
1 2 3
2 4 6

array = [1,10,100]

Result:

A   B   C
1   2   3
2   4   6
10  20  30
20  40  60
100 200 300
200 400 600

Array and dataframe length can be of any size, not necessarily the same.

I have done using FOR loop which is too slow if I have large dataframe or too many of them.

Below is my sample snippet:

for i in array:
    pr[pr.select_dtypes(include=['number']).columns] *= i
    fdf = np.concat([fdf,pr],axis=0)

Is there a much faster way to do this. I am dealing with multiple dataframes which after this operation needs to concatenated.

w-m · Accepted Answer · 2018-08-22 13:22:30Z

3

mult = df.values * np.array([1,10,100])[:, np.newaxis, np.newaxis]
pd.DataFrame(mult.reshape((-1, 3)), columns=df.columns)

     A    B    C
0    1    2    3
1    2    4    6
2   10   20   30
3   20   40   60
4  100  200  300
5  200  400  600

edited Aug 22, 2018 at 13:22

answered Aug 22, 2018 at 13:18

w-m

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Comments

piRSquared · Accepted Answer · 2018-08-22 13:17:26Z

2

Use pandas.concat

pd.concat([df * n for n in a], ignore_index=True)

     A    B    C
0    1    2    3
1    2    4    6
2   10   20   30
3   20   40   60
4  100  200  300
5  200  400  600

Setup

io_ = pd.io.common.StringIO

def rpd(text='', sep='\s{1,}', *args, **kwargs):
  kw = dict(engine='python', sep=sep)
  return pd.read_csv(io_(text), *args, **kw, **kwargs)

df = rpd("""\
A B C
1 2 3
2 4 6""")

a = np.array([1, 10, 100])

edited Aug 22, 2018 at 13:17

answered Aug 22, 2018 at 13:14

piRSquared

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2 Comments

jezrael Over a year ago

ignore_index=True ?

piRSquared Over a year ago

Yes (-: was getting there

BENY · Accepted Answer · 2018-08-22 13:36:06Z

2

IIUC

pd.DataFrame(np.vstack([df.values*x for x in ary]))
Out[171]: 
     0    1    2
0    1    2    3
1    2    4    6
2   10   20   30
3   20   40   60
4  100  200  300
5  200  400  600

pandas reindex

df.reindex(df.index.tolist()*(len(ary))).reset_index(drop=True).mul(pd.Series(np.repeat(ary,len(df))),0)
Out[201]: 
     A    B    C
0    1    2    3
1    2    4    6
2   10   20   30
3   20   40   60
4  100  200  300
5  200  400  600

edited Aug 22, 2018 at 13:36

answered Aug 22, 2018 at 13:20

BENY

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Comments

jadki · Accepted Answer · 2018-08-22 13:18:21Z

0

df = pd.DataFrame({'A':[1,2],'B':[2,4],'C':[3,6]})
ar = [1,10,100]

result = pd.concat([df * i for i in ar], ignore_index = True)

answered Aug 22, 2018 at 13:18

jadki

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1 Comment

piRSquared Over a year ago

This is exactly my answer

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