I have a numpy array that looks like this: [1 -1 -1 1 1 1 -1 -1 -1 1 -1 -1 1 -1]
How do i find the location of the sequence [1 1 1 -1] ?
The output of the function should be something like: Occurrence = 3, since the sequence starts at index 3.
Thanks for helping me!
In one line :
[i for i in range(0,len(x)) if list(x[i:i+4])==[1, 1, 1, -1]]
[3]
IF you want a general solution:
#define your np.array and your list
x=np.array([1 ,-1, -1 ,1 ,1 ,1, -1, -1, -1 ,1, -1, -1, 1 ,-1])
sublist=[1, 1, 1, -1]
[i for i in range(0,len(x)) if list(x[i:i+len(sublist)])==sublist]
[3]
If you have other range of numbers, a double loop will work ...
def subindex(sub, arr):
index = i = -1 # or None or whatever is relevant when sub is NOT found
ext_arr = list(arr)
ext_arr.extend([np.NaN]*len(sub))
for j, sub_j in enumerate(sub):
for i, arr_i in enumerate(ext_arr[j:j+len(sub)]):
try:
if arr_i != sub_j:
continue # get out of the inner loop and check from next i
except:
pass
# if you are here, you have a match
index = i
break
return index
Result ...
>>> arr = "1 -1 -1 1 1 1 -1 -1 -1 1 -1 -1 1 -1".split(" ")
>>> sub = "1 1 1 -1".split()
>>> print(subindex(sub, arr))
3
timeitto check for yourself).