I am currently using a nested for loop to iterate through to arrays to find values that match a certain criterion. The problem is that this method is incredibly inefficient and time consuming. I was told that a better way might be to sort the two arrays based on the data, but this requires me to combine several 1D arrays and one multi-D array, sort based on one column, then separate them again. Is there a more efficient way of doing this? Here is a sample of my code:

x1 = []
x2 = []
velocity = []
plane1Times = np.array([[2293902],[2848853],[482957]])
plane2Times = np.array([[7416504],[2613113],[2326542]])
plane1Local = np.array([[0,0,0],[0,u,0],[0,2*u,0],[u,0,0],[u,u,0],[u,2*u,0],[2*u,0,0],[2*u,u,0],[2*u,2*u,0],[3*u,0,0],[3*u,u,0],[3*u,2*u,0]],dtype='float')
plane2Local = np.array([[0,0,D],[0,u,D],[0,2*u,D],[u,0,D],[u,u,D],[u,2*u,D],[2*u,0,D],[2*u,u,D],[2*u,2*u,D],[3*u,0,D],[3*u,u,D],[3*u,2*u,D]],dtype='float')
for i in range(0,len(plane1Times)):tic = time.time()for n in range(0,len(plane2Times)):if plane2Times[n] - plane1Times[i] <= 10000 and plane2Times[n] - plane1Times[i] > 0:x1 = plane1Local[plane1Dets[i]]x2 = plane2Local[plane2DetScale[n]]distance = np.sqrt((x2[0]-x1[0])**2 + (x2[1]-x1[1])**2 + (x2[2])**2)timeSeparation = (plane2Times[n]-plane1Times[i])*timeScalevelocity += distance/timeSeparationbreak

To give you an example of the time it is currently taking, each array of times is 10**6 values long so 100 loops in i takes about 60 seconds. Can someone please help me?

I cant really test because the code you provided isn't complete, but this is a possible solution

for index,value in enumerate(plane1Times):vec = plane2Times - valuerow,col = np.where((vec<=10000)&(vec>0))if len(row) > 0:x1 = plane1Local[plane1Dets[index]]x2 = plane2Local[plane2DetScale[row[0]]]distance = np.sqrt((x2[0] - x1[0]) ** 2 + (x2[1] - x1[1]) ** 2 + (x2[2]) ** 2)timeSeparation = (plane2Times[row[0]] - plane1Times[index]) * timeScalevelocity += distance / timeSeparation

Eliminate the second loop, and just do the subtraction all at once. Then search the new array, where it meats your criteria. Since it seems that you want the first value, just take the first index like row[0] to get the index of the value check. Removing the second for loop should drop the time considerably.