pandas dataframe multiply with a series [duplicate

Question

This question already has an answer here:

• How do I operate on a DataFrame with a Series for every column 2 answers

What is the best way to multiply all the columns of a Pandas DataFrame by a column vector stored in a Series? I used to do this in Matlab with repmat(), which doesn't exist in Pandas. I can use np.tile(), but it looks ugly to convert the data structure back and forth each time.

Thanks.

Answer 1

This can be accomplished quite simply with the DataFrame method apply.

In[1]: import pandas as pd; import numpy as np

In[2]: df = pd.DataFrame(np.arange(40.).reshape((8, 5)), columns=list('abcde')); df
Out[2]: 
        a   b   c   d   e
    0   0   1   2   3   4
    1   5   6   7   8   9
    2  10  11  12  13  14
    3  15  16  17  18  19
    4  20  21  22  23  24
    5  25  26  27  28  29
    6  30  31  32  33  34
    7  35  36  37  38  39

In[3]: ser = pd.Series(np.arange(8) * 10); ser
Out[3]: 
    0     0
    1    10
    2    20
    3    30
    4    40
    5    50
    6    60
    7    70

Now that we have our DataFrame and Series we need a function to pass to apply.

In[4]: func = lambda x: np.asarray(x) * np.asarray(ser)

We can pass this to df.apply and we are good to go

In[5]: df.apply(func)
Out[5]:
          a     b     c     d     e
    0     0     0     0     0     0
    1    50    60    70    80    90
    2   200   220   240   260   280
    3   450   480   510   540   570
    4   800   840   880   920   960
    5  1250  1300  1350  1400  1450
    6  1800  1860  1920  1980  2040
    7  2450  2520  2590  2660  2730

df.apply acts column-wise by default, but it can can also act row-wise by passing axis=1 as an argument to apply.

In[6]: ser2 = pd.Series(np.arange(5) *5); ser2
Out[6]: 
    0     0
    1     5
    2    10
    3    15
    4    20

In[7]: func2 = lambda x: np.asarray(x) * np.asarray(ser2)

In[8]: df.apply(func2, axis=1)
Out[8]: 
       a    b    c    d    e
    0  0    5   20   45   80
    1  0   30   70  120  180
    2  0   55  120  195  280
    3  0   80  170  270  380
    4  0  105  220  345  480
    5  0  130  270  420  580
    6  0  155  320  495  680
    7  0  180  370  570  780

This could be done more concisely by defining the anonymous function inside apply

In[9]: df.apply(lambda x: np.asarray(x) * np.asarray(ser))
Out[9]: 
          a     b     c     d     e
    0     0     0     0     0     0
    1    50    60    70    80    90
    2   200   220   240   260   280
    3   450   480   510   540   570
    4   800   840   880   920   960
    5  1250  1300  1350  1400  1450
    6  1800  1860  1920  1980  2040
    7  2450  2520  2590  2660  2730

In[10]: df.apply(lambda x: np.asarray(x) * np.asarray(ser2), axis=1)
Out[10]:
       a    b    c    d    e
    0  0    5   20   45   80
    1  0   30   70  120  180
    2  0   55  120  195  280
    3  0   80  170  270  380
    4  0  105  220  345  480
    5  0  130  270  420  580
    6  0  155  320  495  680
    7  0  180  370  570  780

Answer 2

What's wrong with

result = dataframe.mul(series, axis=0)

?

http://pandas.pydata.org/pandas-docs/stable/basics.html#flexible-binary-operations

Answer 3

Why not create your own dataframe tile function:

def tile_df(df, n, m):
    dfn = df.T
    for _ in range(1, m):
        dfn = dfn.append(df.T, ignore_index=True)
    dfm = dfn.T
    for _ in range(1, n):
        dfm = dfm.append(dfn.T, ignore_index=True)
    return dfm

Example:

df = pandas.DataFrame([[1,2],[3,4]])
tile_df(df, 2, 3)
#    0  1  2  3  4  5
# 0  1  2  1  2  1  2
# 1  3  4  3  4  3  4
# 2  1  2  1  2  1  2
# 3  3  4  3  4  3  4

However, the docs note: "DataFrame is not intended to be a drop-in replacement for ndarray as its indexing semantics are quite different in places from a matrix." Which presumably should be interpreted as "use numpy if you are doing lots of matrix stuff".