[英]How to count occurrence of each unique ID in each bin in a 2D histogram (python or pandas)
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.我有一个 csv 文件,我想创建一个二维直方图,其中每个 bin 中的值取决于唯一 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.
例如(见下文),对于范围 0<x<1 和 1<y<2,值是 2 (A, B) 而不是 3 (A, A, B),因为 A 出现了两次。 Thanks!
谢谢!
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 ![]() |
A bin of i_x < x < j_x
, i_y < y < j_y
can be uniquely identified as the (i_x, i_y)
; i_x < x < j_x
, i_y < y < j_y
的 bin 可以唯一标识为(i_x, i_y)
; we can see that this tuple is unique for each bin.我们可以看到这个元组对于每个 bin 都是唯一的。
i_x
and i_y
are simply the floor value of x
and y
. i_x
和i_y
只是x
和y
的底值。 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)
.就像对于行:
(x, y) = (0.5, 1.4)
bin 是: 0 < 0.5 < 1
, 1 < 1.4 < 1.2
这里i_x = 0 = floor(0.5)
和i_y = 1 = floor(1.4)
。
Approach:方法:
i_x
and i_y
for x and y columns.i_x
和i_y
。(i_x, i_y)
and count unique IDs
in each of the group.(i_x, i_y)
对 dataframe 进行分组,并计算每个组中的唯一IDs
。 Code:代码:
>>> 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.如果您将直方图定义为大小为 (5, 5) 的 numpy 数组,那么我们可以将
cnt
值分配给该数组并获得所需的直方图。
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.]])
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