Your task: Define build$\_$cluster$\_$matrix as follows:

Given a k$-$means clustering result, construct a k$-$means assignment assignment matrix. From the preceeding discussion, it is R matrix.

Inputs:

$*)$ labels: A $1-$D array where labels$[$i$]$ is the cluster ID assigned to document i

$*)$max$\_$label: The largest cluster ID plus $1.$ That is$,0<=$ labels$[$i$]<$ max$\_$label for all i

Return: R$,$ a scipy COO matrix

Steps:

The returned matrix R should have len$($labels$)$ rows and max$\_$label columns. It should only have entries equal to $1\mathrm{.}0$ at positions $($i$,$j$)$ where j $==$ labels$[$i$]$; all other entries should be regarded as zero values and therefore not stored explicitly.

Example:

Suppose we run:

build$\_$cluster$\_$matrix$($np$.$array$([1,2,0,0,2,2,2,3,0,3]),4)$

A correct implementation would return:

$(0,1)1\mathrm{.}0$

$(1,2)1\mathrm{.}0$

$(2,0)1\mathrm{.}0$

$(3,0)1\mathrm{.}0$

$....$

$(9,3)1\mathrm{.}0$

Solution:

def build$\_$cluster$\_$matrix$($labels: np$.$ndarray, max$\_$label: int$)->$ sp$.$sparse.coo$\_$matrix:

##code here

demo$\_$R $=$ build$\_$cluster$\_$matrix$($np$.$array$([1,2,0,0,2,2,2,3,0,3]),4)$

print$($demo$\_$R$)$