Kmeans with groupby in dataframe and get cluster in python-CodePudding

I am working with a DataFrame like this:

df=pd.DataFrame({'ID':['12345','55689','56964','49649','89645','0001',
                       '033','03330','064963','306193','03661','1666'],
                 'Culture':['A','A','A','A','A','A','B','B','B','B','B','B'],
                 'H': [30,42,25,32,12,10,4,6,5,10,24,21],
                 'S':[10,76,100,23,65,94,67,24,67,54,87,81],
                 'mean': [23,78,95,52,60,76,68,92,34,76,34,12]})

And first I selected just one group by df_1=df.loc[(df['Culture']=='A') to do kmeans like this

m=df_1.loc[:,['H','mean','mean']].to_numpy()
km = KMeans(n_clusters=3, init='random', max_iter=100, n_init=1, verbose=1)
kmeans_predict = km.predict(m)

array([0, 2, 1, 1, 0, 0], dtype=int32)

clusters = {}
n = 0
for item in kmeans_predict:
  if item in clusters:
    clusters[item].append(list_x1[n])
  else:
    clusters[item] = [list_x1[n]]
  n  =1

And I got something like this after more code:

ID    Culture S mean Cluster
12345  A    10   23    0 
55689  A    76   78   2
56964  A    100   95   1
49649  A    23   52   1
89645  A    65    60   0
00001  A    94    92   0

My goal is do kmeans to every group in this dataframe, but I do not want to do all this group by group (Culture, because there are more than 75 groups). I tried something like:

def cluster(X):
     k_means = KMeans(n_clusters=3).fit(m).groupby('CUL')
     X['cluster'] = k_means.labels_
     return X

 df= cities_e.groupby('CUL').apply(cluster)

Trying to have all this clustering inside each group by 'Culture' and get it's predicted cluster in the DataFrame.

CodePudding user response：

You could simply wrap your code in a function and use groupby.apply. However, to get the indexes return a Series, instead of an array:

from sklearn.cluster import KMeans
def get_cluster(df_1):
    m=df_1.loc[:,['H','mean','mean']].to_numpy()
    km = KMeans(n_clusters=3, init='random', max_iter=100, n_init=1, verbose=0).fit(m)
    kmeans_predict = km.predict(m)
    return pd.Series(kmeans_predict, index=df_1.index)

df['Cluster'] = df.groupby('Culture').apply(get_cluster).droplevel(0)

Output:

        ID Culture   H    S  mean  Cluster
0    12345       A  30   10    23        2
1    55689       A  42   76    78        0
2    56964       A  25  100    95        1
3    49649       A  32   23    52        2
4    89645       A  12   65    60        2
5     0001       A  10   94    76        1
6      033       B   4   67    68        1
7    03330       B   6   24    92        0
8   064963       B   5   67    34        2
9   306193       B  10   54    76        0
10   03661       B  24   87    34        2
11    1666       B  21   81    12        2

If you want distinct cluster number across different Cultures, we could assign a group number for each Culture, then use it to modify cluster numbers:

def get_cluster(df_1):
    m=df_1.loc[:,['H','mean','mean']].to_numpy()
    km = KMeans(n_clusters=3, init='random', max_iter=100, n_init=1, verbose=0).fit(m)
    kmeans_predict = km.predict(m)   3 * df_1['Culture_id'].iat[0]
    return pd.Series(kmeans_predict, index=df_1.index)

g = df.groupby('Culture')
df['Culture_id'] = g.ngroup()
df['Cluster'] = g.apply(get_cluster).droplevel(0)
df = df.drop(columns=['Culture_id'])

Output:

        ID Culture   H    S  mean  Cluster
0    12345       A  30   10    23        0
1    55689       A  42   76    78        1
2    56964       A  25  100    95        1
3    49649       A  32   23    52        0
4    89645       A  12   65    60        2
5     0001       A  10   94    76        2
6      033       B   4   67    68        3
7    03330       B   6   24    92        5
8   064963       B   5   67    34        4
9   306193       B  10   54    76        3
10   03661       B  24   87    34        4
11    1666       B  21   81    12        4