Replace row of a pandas dataframe with numpy array

Question

I have a pandas dataframe, for example:

import sys
if sys.version_info[0] < 3: 
    from StringIO import StringIO
else:
    from io import StringIO

import pandas as pd

TESTDATA = StringIO("""col1;col2;col3
    (1,2);(3,4);(5,6)
    (7,8);(9,10);(11,12)
    (13,14);(15,16);(17,18)
    (19,20);(21,22);(23,24)
    """)

df = pd.read_csv(TESTDATA, sep=";")

dataframe looks like this:

          col1     col2     col3
0        (1,2)    (3,4)    (5,6)
1        (7,8)   (9,10)  (11,12)
2      (13,14)  (15,16)  (17,18)
3      (19,20)  (21,22)  (23,24)

Now, let's say there's 2 rows 0 and 2 that I wish to replace with a numpy array of zeros such as:

np.zeros((1,3,2))
"""array([[[ 0.,  0.],
        [ 0.,  0.],
        [ 0.,  0.]]])"""

So the final result is that for the rows 0 and 2, I have tuples (0,0), (0,0), and (0,0). Could someone propose an approach?

How do you connect 6 values in np.zeros((1,3,2)) with 6 values in a row? There's more than one way you can align the 2 structures (flatten by row, by column, etc). — jpp
– jpp, Commented Mar 22, 2018 at 11:09

IanS · Accepted Answer · 2018-03-22 14:19:15Z

1

This would work:

arr = np.zeros((1, 3, 2))
row = [tuple(t) for t in arr[0]]
df.loc[0] = row
df.loc[2] = row

answered Mar 22, 2018 at 14:19

IanS

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