finding closest values in two arrays without repetitions

Question

I have 2 arrays list1 and list2

list1 = np.array([1052, 1092, 1117,1135, 1156,1212,1249,1325,1363,1380,1382,1400,1484])
list2 = np.array([1053.435, 1093.698, 1118.342, 1136.697,
                  1157.832, 1212.938, 1215.741, 1250.266, 1252.352,
                  1254.453, 1256.831, 1259.182, 1326, 1367.228,
                  1400, 1410, 1740])

for each element in list1 I would like to find the closest element in list2 and return its index.

mylist = []
for idxlabel in range(0,len(list1)): 
    a = min(enumerate(list2), key=lambda x:abs(x[1]-list1[idxlabel]))
    print a
    mylist.append(np.copy(a))

My problem is that after that one element in list2 is found as "best match" I would like to remove it from the search to avoid that different element in list1 match with the same element in list2

(0, 1053.4349999999999)
(1, 1093.6980000000001)
(2, 1118.3420000000001)
(3, 1136.6969999999999)
(4, 1157.8320000000001)
(5, 1212.9380000000001)
(7, 1250.2660000000001)
(12, 1326.0)
(13, 1367.2280000000001)
(13, 1367.2280000000001)
(13, 1367.2280000000001)
(14, 1400.0)
(15, 1410.0)

In this example indeed the the 9th 10th and 11th elements of list1 match all with the 13th element of list2 and this is NOT desired...

@Mr.Xcoder yes sure, this is just a piece of code I found to find the closest match... feel free to do it in another way — gabboshow, Mar 23 '17 at 14:38
`np.searchsorted(list2, list1)` is probably a good starting point (if indeed they are sorted) — Jon Clements, Mar 23 '17 at 14:54
Hi @JonClements I get as well repeated index with np.serachsorted — gabboshow, Mar 23 '17 at 15:01
@Divakar the question you pointed me does not answer my question... it also replicates the matches — gabboshow, Mar 23 '17 at 16:24
@gabboshow Could you add the expected output into the question? — Divakar, Mar 23 '17 at 16:47

score 0 · Answer 1 · answered Mar 23 '17 at 15:05

Probably not the best way to do it, but you could keep a copy of that list2 array and modify it every time you find a match. Something like this:

list2aux = list(list2)
mylist = []
for idxlabel in range(0,len(list1)): 
    a = min(enumerate(list2aux), key=lambda x:abs(x[1]-list1[idxlabel]))
    list2aux[a[0]] = 0
    print(a)
    mylist.append(np.copy(a))

Changing that element in list2aux to 0 will make the values that were already selected as best matches not to be selected again.

finding closest values in two arrays without repetitions

1 Answers1