Formating values in between of arrays Numpy Python

Question

I am trying to code a function where it checks if the Numbers elements are between the Formating values. So for the first element of Numbers 3 is not between the first and the second element 0, 2 of Formating so it returns 0 as the result. For the elements 3,4 in Numbers are between 2 ,5 so it return those 3,4 as the numbers between 2,5. The index value keeps on getting incremented until the end of the Formating array. How could I fix the function below so it gives the expected output, without a for loop.

Numbers = np.array([3, 4, 5, 7, 8, 10,20])
Formating = np.array([0, 2 , 5, 12, 15, 22])

#previouse and curent index values 
prevs = np.arange(0,len(Formating)-1,1)
currents = np.arange(1,len(Formating),1)
index = 0

np.where(Formating[prevs[index]] <= Numbers < Formating[currents[index]], Numbers, ++index)

Expected Output:

Numbers between 0,2 = 0
Numbers between 2,5 = 3,4
Numbers between 5,12 = 5,7,8,10
Numbers between 12,15 = 0
Numbers between 15,22 = 20

Something with np.histogram(Numbers, bins=Formating) or np.digitize(Numbers, Formating)? — Quang Hoang
– Quang Hoang, Commented Mar 10, 2021 at 19:23
Usually formatting refers to converting numbers to strings, as in %d%3` or {}.format(3). Your use of Formating as an array, as though it means something, is misleading, and may discourage useful answers. Numbers isn't that great of a variable either (it looks more like a class name). — hpaulj
– hpaulj, Commented Mar 10, 2021 at 21:00

Bob · Accepted Answer · 2021-03-10 21:31:32Z

0

You could sort then search sorted and it will run more efficiently (except for the IO)


Numbers = np.array([3, 4, 5, 7, 8, 10,20])
Formating = np.array([0, 2 , 5, 12, 15, 22])
x = np.sort(Numbers);
l = np.searchsorted(x, Formating, side='left')

for i in range(len(l)-1):
    if l[i] >= l[i+1]:
        print('Numbers between %d,%d = _0_' % (Formating[i], Formating[i+1]))
    else:
        print('Numbers between %d,%d = %s' % (Formating[i], Formating[i+1], ','.join(map(str, list(x[l[i]:l[i+1]])))))

answered Mar 10, 2021 at 21:31

Bob

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