Pandas Dataframe Update Rows

Question

Let's say that I have a dataframe, X, initiated with 0s and a dimension m x n. I have n unique values (1,2,3,...,n) in a pandas.series, Y, that has length m. How do I set the Y[i] column of the ith row of X (change 0 to 1) efficiently without using a loop. Especially for large m and n.

For example, for Y = [3,2,1]
X
row     1       2      3
0       0       0      0
1       0       0      0
2       0       0      0

to
row     1       2      3
0       0       0      1
1       0       1      0
2       1       0      0

What is the issue, exactly? Have you tried anything, done any research? Also, why are you seemingly using 0 and 1 instead of actual boolean values? — AMC
– AMC, Commented Feb 19, 2020 at 23:12
If your matrix is not square (i.e. I assume m does not necessarily equal n), then it is not helpful to have a square matrix as your example. — Alexander
– Alexander, Commented Feb 19, 2020 at 23:21
Sheer curiosity. I wanted to know if there was a built-in function for something like that. Instead of using loop and .iloc. — Amartya Barua
– Amartya Barua, Commented Feb 20, 2020 at 0:32
Note that iat is faster for setting scalar values compared to iloc. — Alexander
– Alexander, Commented Feb 20, 2020 at 1:06

Alexander · Accepted Answer · 2020-02-20 01:03:43Z

2

I'm not sure why you are against for loops. This should be fairly efficient.

for row, col in enumerate(Y):
    df.iat[n, col] = 1

You could also compute the index locations and set their values to one, then reshape the result to the m x n shape of the matrix.

Y = [3, 2, 1]
n = 5
m = len(Y)
locations = set(row * n + col for row, col in enumerate(Y))
df = pd.DataFrame(
    np.array([1 if idx in locations else 0 for idx in range(m * n)]).reshape((m, n))
)
>>> df
   0  1  2  3  4
0  0  0  0  1  0
1  0  0  1  0  0
2  0  1  0  0  0

edited Feb 20, 2020 at 1:03

answered Feb 19, 2020 at 23:26

Alexander

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