Sum based on a criterion

Question

I have a data sheet where I want to do a calculation until I find a zero in one specific column. Then I want to sum all the results of this calculation up to that zero and save the result in an array. I tried with an np array and a list but it does not work:

tst = []
x = data[1:len(data),0]
y = data[1:len(data),1]  
intt = data[1:len(data),2]
for i in range(0,len(data)):
   if intt[i]!=0:         
      tst.append((x[i]**2.0+ y[i]-y[i-1])**2.0)

I want the tst to contain in each position, the sum of the expression in the append().

Thanks!

Just add else: break. But also think about what y[i-1] is when i is 0. — trincot
– trincot, Commented Jan 17 at 21:26

Slavensky · Accepted Answer · 2025-01-17 22:42:20Z

1

I looks like you skip the first datapoint since you write x = data[1:len(data),0] and so forth. As @trincot mentioned, you also have to care about the y[i-1] case for i=0. Maybe the following will help you:

tst = []
x = data[:,0]
y = data[:,1]  
intt = data[:,2]
total = []
for i in range(1,len(data)):
   if intt[i]!=0:         
      tst.append((x[i]**2.0+ y[i]-y[i-1])**2.0)
   else:
      total.append(sum(tst))
      break
total = np.array(total)

This includes all data points in x, y, and intt, but the first data point will still be skipped since the loop starts with i=1.

Edit: A cleaner solution, based on comment from @Barmar.

tst = []
x = data[:,0]
y = data[:,1]  
intt = data[:,2]
for i in range(1,len(data)):
   if intt[i]!=0:         
      tst.append((x[i]**2.0+ y[i]-y[i-1])**2.0)
   else:
      total = np.array(sum(tst))
      break

Edit: Another solution that goes through all of intt:

tst = []
x = data[:,0]
y = data[:,1]  
intt = data[:,2]
total = []
for i in range(1,len(data)):
   if intt[i]!=0:         
      tst.append((x[i]**2.0+ y[i]-y[i-1])**2.0)
   elif i == len(data) - 1:
      total.append(sum(tst))
   else:
      total.append(sum(tst))
      tst = []
      continue
total = np.array(total)

edited Jan 17 at 22:42

answered Jan 17 at 21:56

Slavensky

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6 Comments

Riri Jan 17 at 22:07

The problem is that append only appends the result of the calculation. I want to have the sum of all these calculations while the if statement is True.

Slavensky Jan 17 at 22:10

Do you still ned the tst list then? I have given an update to my answer where I have added a total llist, which simply appends the sum of the tst array when a 0 is encountered.

Barmar Jan 17 at 22:18

Why is total a list? You only append to it once.

Slavensky Jan 17 at 22:22

@Riri asked for it to be an array. Of course, a np.array(total) at the end of the script will turn it into a numpy array. One could just let total be equal to a numpy array of the sum of tst. I have added that possibilty too in my answer now.

Riri Jan 17 at 22:29

@Slavensky I think I don't explain it very well: When I find a zero I want to add up all the numbers I have appended to tst up to that point. And save this sum in a new list. Hope that makes it clearer.

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