Remove values from numpy array closer to each other

In the following is provided an additional solution using numpy functionalities (more precisely np.ediff1d which makes the differences between consecutive elements of a given array. This code considers as threshold the value associated to the th variable.

a = np.array([1,2,10,11,18,19])
th = 1
b = np.delete(a, np.argwhere(np.ediff1d(a) <= th) + 1) # [1, 10, 18]

Here is simple function to find the first values of series of consecutives values in a 1D numpy array.

import numpy as np

def find_consec(a, step=1):
    vals = []
    for i, x in enumerate(a):
        if i == 0:
            diff = a[i + 1] - x
            if diff == step:
                vals.append(x)
        elif i < a.size-1:
            diff = a[i + 1] - x
            if diff > step:
                vals.append(a[i + 1])
    return np.array(vals)

a = np.array([1,2,10,11,18,19])
find_consec(a) # [1, 10, 18]

Welcome to stackoverflow. below is the code that can answer you question:

def closer(arr,cozy):
    result = []
    result.append(arr[0])
    for i in range(1,len(arr)-1):
        if arr[i]-result[-1]>cozy:
            result.append(arr[i])
    print result           

Example:

a = [6,10,7,20,21,16,14,3,2]
a.sort()
closer(a,1)
output : [2, 6, 10, 14, 16, 20]
closer(a,3)
Output: [2, 6, 10, 14, 20]

Sorry for being late to the party. But Agglomerative Clustering may be what you're looking for. By specifying the distance threshold and linkage criterion, the algorithm will return the desired result.

from sklearn.cluster import AgglomerativeClustering
a = np.array([1,2,10,11,18,19])
dups_filter = AgglomerativeClustering(linkage="single", distance_threshold=1.1, metric="manhattan", n_clusters=None)

cluster_labels = dups_filter.fit_predict(a.reshape(-1,1))
filtered_array = np.array([])
for c in range(dups_filter.n_clusters_):
 filtered_array = np.append(filtered_array, np.min(a[cluster_labels==c]))