Sum of consecutive data in List - Python - python-3.x

Question

I have a long list of number of which a sample look something like shown below:
L = [0, 1, 2, 0, 0, 2, 3, 1, 0, 10, 0, 2, 4, 5, 0, 0, 7, 8, 9, 0, ...]
singleData = []
sumofTwo = []
sumofThree = []
sumofFour = []
.
.
What I want to be able to do is categorize a data OR sum of two or more consecutive data into the respective lists based on the COUNT of numbers involved in the sum operation. So, if there is an occurrence of zero between two numbers, then their sum would not be considered.
For example, if I take the list above, the sum of 1 and 2 is to be added to the list sumofTwo. Similarly, if it is the sum of three consecutive numbers without the occurrence of 0's then the sum is expected to be in the sumofThree list (sum of 2,3,1; 2,4,5 and 7, 8, 9). If a number occurs between two or more 0's then it is to be appended to the singleData list (eg. 10).
How can i achieve this considering that in the list(L) there can be a sum of random consecutive numbers? Example, sum of 6 or 7 or 8 or any consecutive numbers?
I was able to segregate only a single number between 0's and the sum of two numbers. Following is the code:
for i in range(len(l)):
try:
if i == 0:
if l[i] == 0:
continue
elif l[i] != 0 and l[i+1] == 0:
singleData.append(l[i])
elif l[i] != 0 and l[i+1] != 0:
sumofTwo.append(l[i]+l[i+1])
elif i == len(l)-1:
if l[i] != 0 and l[i-1] == 0:
singleData.append(l[i])
else:
if l[i] != 0:
if l[i+1] == 0 and l[i-1] == 0:
singleData.append(l[i])
elif l[i+1] != 0:
sumofTwo.append(l[i]+l[i+1])
except IndexError:
print("Index out of range")
I realized that my code will only get messier with more cases of the sum of consecutive numbers and ultimately end up with error.
Can anybody help me out? Thanks in advance :))))

Answer 1

I would recommend using a dictionary to store the results:
L = [0, 1, 2, 0, 0, 2, 3, 1, 0, 10, 0, 2, 4, 5, 0, 0, 7, 8, 9, 0]
consecutive_sums = {} # Create an empty dictionary
I would also recommend directly iterating through the list, rather than using range(len(L). e.g.
for number in L:
print(number)
Then, you can just create a counter variable to check how long the current sequence is, and reset the counter when you get to a zero.
L = [0, 1, 2, 0, 0, 2, 3, 1, 0, 10, 0, 2, 4, 5, 0, 0, 7, 8, 9, 0]
consecutive_sums = {} # Create an empty dictionary
counter = 0
sum = 0
for number in L:
if number == 0: # Save the current sum and reset counter
# Check if we've already had a sequence of this length.
# If so, we have already added a list, so can append to it.
if counter in consecutive_sums:
consecutive_sums[counter].append(sum)
else: # If this is the first time we've had this length sequence
# we need to create a new key value pair in the dictionary.
# Note that the value is a list containing `sum`.
# Make sure to use a list so that you can append to it later.
consecutive_sums[counter] = [sum]
# Reset counter and sum
counter = 0
sum = 0
else: # Increment counter
counter += 1
sum += number
print(consecutive_sums)
Note that this code will not sort the dictionary keys, so the sums of sequences of length 1 may not appear at the beginning. But you can access it using consecutive_sums[1].
Also note that this version of the code also counts sequences of length 0. I suspect this is not what you want, but I'll let you figure out how to fix it!
Output:
{0: [0, 0, 0], 2: [3], 3: [6, 11, 24], 1: [10]}
EDIT:
I've intentionally tried to solve this using just builtin functions and datatypes. But if you really want to be fancy, you can use collections.defaultdict.
Below is an alternative version which uses defaultdict:
from collections import defaultdict
L = [0, 1, 2, 0, 0, 2, 3, 1, 0, 10, 0, 2, 4, 5, 0, 0, 7, 8, 9, 0]
consecutive_sums = defaultdict(list) # Create an empty defaultdict of type list
counter = 0
sum = 0
for number in L:
if number == 0: # Save the current sum and reset counter
# The magic of defaultdict:
# We don't need to check if the key exists.
# If it doesn't exist yet, defaultdict will automatically make
# an empty list for us to append to!
consecutive_sums[counter].append(sum)
# Reset counter and sum
counter = 0
sum = 0
else: # Increment counter
counter += 1
sum += number
print(consecutive_sums)

Answer 2

The first thing I would do is to organize a bit better our output lists, so that each can be accessed in the same way using the number of consecutive numbers. As #daviewales suggested, you could do this with a dictionary with lists as values, something like sums = {}, so that sums[1] would be the same as singleData, sums[2] the same as sumofTwo, and so on. That way, you will avoid a lot of if to know in what list you should put your data, you'll just need to use stuff like sums[nbOfValuesInSum].
Second thing, you could write a function that detects your sequences of non-zero values. A possibility would be a function that takes a list and a start index, and returns the start and end indexes of the next "interesting" sequence. It would look like this :
def findNextSequence(l, start):
while l[start] == 0:
if start == len(l)-1:
return None # there is no non-zero value left
start+=1
# when we exit the loop, start is the index of the first non-zero value
end = start + 1
while l[end] != 0:
if end == len(l)-1:
break
end+=1
# and now end is the index of the first zero value after the sequence
return (start, end)
Then you can call it in a loop, like this:
i = 0
while True:
bounds = findNextSequence(l, i)
if bounds is None:
break # there is no non-zero value left
seq = l[bounds[0]:bounds[1]] # get the start and end index of the sequence
if len(seq) not in sums:
sums[len(seq)] = []
sums[len(seq)].append(sum(seq)) # see? No need to explicitly check len(seq) to know what list I want
i = bounds[1] # get ready for the next iteration
if i == len(l):
break
NB : no need to pass l as a parameter of findNextSequence if it's a global variable