Find groups of values that are !=0 in a list

Question

I’m looking for an easy way to find "plateaus" or groups in python lists. As input, I have something like this:

mydata = [0.0, 0.0, 0.0, 0.0, 0.0, 0.143, 0.0, 0.22, 0.135, 0.44, 0.1, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.33, 0.65, 0.22, 0.0, 0.0, 0.0, 0.0, 0.0]

I want to extract the middle position of every "group". Group is defined in this case as data that is !=0 and for example at least 3 positions long. Enclaved single zeros (like on position 6) should be ignored.

Basically, I want to get the following output:

myoutput = [8, 20]

For my use case, it is not really important to get very precise output data. [10,21] would still be fine.

To conclude everything: first group: [0.143, 0.0, 0.22, 0.135, 0.44, 0.1]; second group: [0.33, 0.65, 0.22]. Now, the position of the middle element (or left or right from the middle, if there is no true middle value). So in the output 8 would be the middle of the first group and 20 the middle of the second group.

I’ve already tried some approaches. But they are not as stable as I wanted them to be (for example: more enclaved zeros can cause problems). So before investing more time in this idea, I wanted to ask if there is a better way to implement this feature. I even think that this could be a generic problem. Is there maybe already standard code that solves it?

There are other questions that describe roughly the same problem, but I have also the need to "smooth" the data before processing.

smooth the data – get rid of enclaved zeros

 import numpy as np
 def smooth(y, box_pts):
     box = np.ones(box_pts)/box_pts
     y_smooth = np.convolve(y, box, mode='same')
     return y_smooth

 y_smooth = smooth(mydata, 20)

find start points in the smooth list (if a value is !=0 and the value before was 0 it should be a start point). If an endpoint was found: use the last start point that was found and the current endpoint to get the middle position of the group and write it to a deque.

 laststart = 0
 lastend = 0
 myoutput = deque()

 for i in range(1, len(y_smooth)-1):
         #detect start:
         if y_smooth[i]!=0 and y_smooth[i-1]==0:
             laststart = i   
         #detect end:
         elif y_smooth[i]!=0 and y_smooth[i+1]==0 and laststart+2 < i:
             lastend = i
             myoutput.appendleft(laststart+(lastend-laststart)/2)

EDIT: to simplify everything, I gave only a short example for my input data at the beginning. This short list actually causes a problematic smoothing output – the whole list will get smoothed, and no zero will be left. actual input data; actual input data after smoothing

Asked By: chrisg

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Source

Answer 1

A fairly simple way of finding groups as you described would be to convert data to a boolean array with ones for data inside groups and 0 for data outside the groups and compute the difference of two consecutive value, this way you’ll have 1 for the start of a group and -1 for the end.

Here’s an example of that :

import numpy as np

mydata = [0.0, 0.0, 0.0, 0.0, 0.0, 0.143, 0.0, 0.22, 0.135, 0.44, 0.1, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.33, 0.65, 0.22, 0.0, 0.0, 0.0, 0.0, 0.0]
arr = np.array(mydata)

mask = (arr!=0).astype(np.int) #array that contains 1 for every non zero value, zero other wise
padded_mask =  np.pad(mask,(1,),"constant") #add a zero at the start and at the end to handle edge cases
edge_mask = padded_mask[1:] - padded_mask[:-1] #diff between a value and the following one 
#if there's a 1 in edge mask it's a group start
#if there's a -1 it's a group stop

#where gives us the index of those starts and stops
starts = np.where(edge_mask == 1)[0]
stops = np.where(edge_mask == -1)[0]
print(starts,stops)

#we format groups and drop groups that are too small
groups = [group for group in zip(starts,stops) if (group[0]+2 < group[1])]


for group in groups:
        print("start,stop : {}  middle : {}".format(group,(group[0]+group[1])/2) )

And the output :

[ 5  7 19] [ 6 11 22]
start,stop : (7, 11)  middle : 9.0
start,stop : (19, 22)  middle : 20.5

Answered By: jadsq

Answer 2

Your smoothed data has no zeros left:

import numpy as np

def smooth(y, box_pts):
    box = np.ones(box_pts)/box_pts
    print(box)
    y_smooth = np.convolve(y, box, mode='same')
    return y_smooth

mydata = [0.0, 0.0, 0.0, 0.0,-0.2, 0.143, 
          0.0, 0.22, 0.135, 0.44, 0.1, 0.0, 
          0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 
          0.33, 0.65, 0.22, 0.0, 0.0, 0.0, 
          0.0, 0.0]

y_smooth = smooth(mydata, 27)
print(y_smooth)

Output:

[ 0.0469   0.0519   0.0519   0.0519   0.0519   0.0519   
  0.0519   0.0519  0.0519   0.0519   0.0684   0.1009   
  0.1119   0.1119   0.1119   0.1119  0.10475  0.10475  
  0.09375  0.087    0.065    0.06     0.06     0.06     
  0.06   0.06     0.06   ]

A way to find it in your original data would be:

def findGroups(data, minGrpSize=1):
  startpos = -1
  endpos = -1
  pospos = []
  for idx,v in enumerate(mydata):
    if v > 0 and startpos == -1:
      startpos = idx
    elif v == 0.0:
      if startpos > -1:
       if idx < (len(mydata)-1) and mydata[idx+1] != 0.0:
         pass # ignore one 0.0 in a run
       else:
         endpos = idx

      if startpos > -1:
        if endpos >-1 or idx == len(mydata)-1: # both set or last one 
          if (endpos - startpos) >= minGrpSize:
              pospos.append((startpos,endpos))
          startpos = -1
          endpos = -1
  return pospos

pos = findGroups(mydata,1)
print(*map(lambda x: sum(x) // len(x), pos))

pos = findGroups(mydata,3)
print(*map(lambda x: sum(x) // len(x), pos))

pos = findGroups(mydata,5)
print(*map(lambda x: sum(x) // len(x), pos))

Output:

8 20
8 20
8

Answered By: Patrick Artner

Answer 3

Part 2 – find the group midpoint:

mydata = [0.0, 0.0, 0.0, 0.0, 0.0, 0.143, 0.0, 0.22, 0.135, 0.44, 0.1, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0,
          0.0, 0.0, 0.33, 0.65, 0.22, 0.0, 0.0, 0.0, 0.0, 0.0]

groups = []
last_start = 0
last_end = 0
in_group = 0

for i in range(1, len(mydata) - 1):
    if not in_group:
        if mydata[i] and not mydata[i - 1]:
            last_start = i
            in_group = 1
    else:  # a group continued.
        if mydata[i]:
            last_end = i
        elif last_end - last_start > 1:  # we have a group i.e. not single non-zero value
            mid_point = (last_end - last_start) + last_start
            groups.append(((last_end - last_start)//2) + last_start)
            last_start, last_end, in_group = (0, 0, 0)
        else:  # it was just a single non-zero.
            last_start, last_end, in_group = (0, 0, 0)

print(groups)

Output:

[8, 20]

Answered By: Nabeel Ahmed

Answer 4

Full numpy solution would be something like this: (not fully optimized)

import numpy as np

input_data = np.array([0.0, 0.0, 0.0, 0.0, 0.0, 0.143,
                       0.0, 0.22, 0.135, 0.44, 0.1, 0.0,
                       0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0,
                       0.33, 0.65, 0.22, 0.0, 0.0, 0.0,
                       0.0, 0.0])

# Find transitions between zero and nonzero
non_zeros = input_data > 0
changes = np.ediff1d(non_zeros, to_begin=not non_zeros[0],
                     to_end=not non_zeros[-1])
change_idxs = np.nonzero(changes)[0]

# Filter out small holes
holes = change_idxs.reshape(change_idxs.size//2, 2)    
hole_sizes = holes[:, 1]-holes[:, 0]
big_holes = holes[hole_sizes > 1]

kept_change_idxs = np.r_[0, big_holes.flatten(), input_data.size]

# Get midpoints of big intervals
intervals = kept_change_idxs.reshape(kept_change_idxs.size//2, 2)
big_intervals = intervals[intervals[:, 1]-intervals[:, 0] >= 3]
print((big_intervals[:, 0]+big_intervals[:, 1])//2)

Answered By: kuppern87

Find groups of values that are !=0 in a list

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