Clustering longitudinal data with multiple variables in R - r

Question

I have a dataset that contains the observations of 30 people and each of them had done 20 experiments. Suppose my data looks like this:
ID trial reaction response prop_1 prop_2
"s1" 1 2.12 0 0.52 0.48
"s1" 2 1.32 1 0.12 0.88
"s1" 3 NA 1 NA NA
"s2" 1 2.33 1 0.65 0.35
"s2" 2 2.56 0 0.43 0.57
"s2" 3 NA 1 NA NA
I want to cluster the participants using these variables. I studied traj, latrend and kml packages but all of them use just one variable to cluster the data. How can I use multiple variables to cluster a longitudinal data like this?
Any simple help or guidance would be appreciated.

Answer 1

Here is one way to do it.
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn.cluster import KMeans
import seaborn as sns; sns.set()
import csv
df = pd.read_csv('C:\\business.csv')
df.dropna(axis=0,how='any',subset=['latitude','longitude'],inplace=True)
K_clusters = range(1,10)
kmeans = [KMeans(n_clusters=i) for i in K_clusters]
Y_axis = df[['latitude']]
X_axis = df[['longitude']]
score = [kmeans[i].fit(Y_axis).score(Y_axis) for i in range(len(kmeans))]# Visualize
plt.plot(K_clusters, score)
plt.xlabel('Number of Clusters')
plt.ylabel('Score')
plt.title('Elbow Curve')
plt.show()
X = df[['longitude', 'latitude']].copy()
kmeans = KMeans(n_clusters = 5, init ='k-means++')
kmeans.fit(X[X.columns[1:2]]) # Compute k-means clustering
X['cluster_label'] = kmeans.fit_predict(X[X.columns[1:3]])
centers = kmeans.cluster_centers_ # Coordinates of cluster centers
labels = kmeans.predict(X[X.columns[1:2]]) # Labels of each point
X.head(10)
X.plot.scatter(x = 'latitude', y = 'longitude', c=labels, s=50, cmap='viridis')
plt.scatter(centers[:, 0], centers[:, 1], c='black', s=200, alpha=0.5)
Here's another idea.
# import necessary modules
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.cluster import KMeans
from collections import Counter
df = pd.read_csv('C:\\properties_2017.csv')
# df.head(10)
df = df.head(10000)
list(df)
df.shape
df.shape
df = df.sample(frac=0.2, replace=True, random_state=1)
df.shape
df = df.fillna(0)
df.isna().sum()
df['regionidzip']=df['regionidzip'].fillna(97000)
df.dropna(axis=0,how='any',subset=['latitude','longitude'],inplace=True)
X=df.loc[:,['latitude','longitude']]
zp=df.regionidzip
id_n=8
kmeans = KMeans(n_clusters=id_n, random_state=0).fit(X)
id_label=kmeans.labels_
#plot result
ptsymb = np.array(['b.','r.','m.','g.','c.','k.','b*','r*','m*','r^']);
plt.figure(figsize=(12,12))
plt.ylabel('Longitude', fontsize=12)
plt.xlabel('Latitude', fontsize=12)
for i in range(id_n):
cluster=np.where(id_label==i)[0]
plt.plot(X.latitude[cluster].values,X.longitude[cluster].values,ptsymb[i])
plt.show()
#revise the clustering based on zipcode
uniq_zp=np.unique(zp)
for i in uniq_zp:
a=np.where(zp==i)[0]
c = Counter(id_label[a])
c.most_common(1)[0][0]
id_label[a]=c.most_common(1)[0][0]
#plot result (revised)
plt.figure(figsize=(12,12))
plt.ylabel('Longitude', fontsize=12)
plt.xlabel('Latitude', fontsize=12)
for i in range(id_n):
cluster=np.where(id_label==i)[0]
plt.plot(X.latitude[cluster].values,X.longitude[cluster].values,ptsymb[i])
plt.show()
https://www.kaggle.com/xxing9703/kmean-clustering-of-latitude-and-longitude?select=zillow_data_dictionary.xlsx
https://www.kaggle.com/c/zillow-prize-1/data
Also, check this out.
https://towardsdatascience.com/clustering-geospatial-data-f0584f0b04ec
https://raw.githubusercontent.com/mdipietro09/DataScience_ArtificialIntelligence_Utils/master/machine_learning/data_stores.csv