How to count occurrence of each unique ID in each bin in a 2D histogram (python or pandas)

Question

I have a csv file and I would like to create a 2d histogram where the value in each bin depends on the unique ID. For example (see below), for the range 0<x<1 and 1<y<2, the value is 2 (A, B) not 3 (A, A, B) because A appears twice. Thanks!

ID	x	y
A	0.5	1.4
A	0.6	1.6
A	1.2	2.2
B	0.7	1.7
C	4.4	3.5
C	3.1	3.7

Answer 1

A bin of i_x < x < j_x , i_y < y < j_y can be uniquely identified as the (i_x, i_y) ; we can see that this tuple is unique for each bin. i_x and i_y are simply the floor value of x and y . Like For row: (x, y) = (0.5, 1.4) bin is: 0 < 0.5 < 1 , 1 < 1.4 < 1.2 here i_x = 0 = floor(0.5) and i_y = 1 = floor(1.4) .

Approach:

1. Find i_x and i_y for x and y columns.
2. Group the dataframe using key (i_x, i_y) and count unique IDs in each of the group.

Code:

>>> df
  ID    x    y
0  A  0.5  1.4
1  A  0.6  1.6
2  A  1.2  2.2
3  B  0.7  1.7
4  C  4.4  3.5
5  C  3.1  3.7

df['bin_x'] = np.floor(df.x).astype(int)
df['bin_y'] = np.floor(df.y).astype(int)
df = (df.groupby(['bin_x', 'bin_y'], as_index = False)['ID']
        .agg({'cnt' : 'nunique'}))


>>> df
   bin_x  bin_y  cnt
0      0      1    2
1      1      2    1
2      3      3    1
3      4      3    1

If you are defining your histogram as numpy array of size (5, 5) then we can assign cnt values to that array and get the desired histogram.

histogram = np.zeros((5, 5))
histogram[df.bin_x, df.bin_y] = df.cnt
>>> histogram
array([[0., 2., 0., 0., 0.],
       [0., 0., 1., 0., 0.],
       [0., 0., 0., 0., 0.],
       [0., 0., 0., 1., 0.],
       [0., 0., 0., 1., 0.]])