# Python sum of number in array, ignoring sections of specific numbers

## Question:

I am very new to Python and have been going through multiple tutorials to get better.

I have straggled with one difficult problem and found a solution. But it feels, works very newbie like. I think that I have tailored it to answer the specific question.

So the question is:

SUMMER OF ’69: Return the sum of the numbers in the array, except ignore sections of numbers starting with a 6 and extending to the next 9 (every 6 will be followed by at least one 9). Return 0 for no numbers.

``````summer_69([1, 3, 5]) --> 9
summer_69([4, 5, 6, 7, 8, 9]) --> 9
summer_69([2, 1, 6, 9, 11]) --> 14
``````

My code to solve this is:

``````def summer_69(arr):
list1 = arr
summ = int()
for i in range(0, len(arr)):
if 6 in list1:
listOfRange = range(list1.index(6), list1.index(9) + 1)
for index in listOfRange:
print(listOfRange)
arr[index] = 0
if 6 != arr[i]:
summ += arr[i]
else:
continue
else:
summ += arr[i]
return summ
``````

It is a very basic problem and I am very alerted that I have struggled with something like this already.

Here’s how I’d do it, as a first cut:

``````def summer_69(series):
in_summer = False
cur_sum = 0
for v in series:
if in_summer:
if v == 9:
in_summer = False
else:
if v == 6:
in_summer = True
else:
cur_sum += v
return cur_sum
``````

Here’s a version that uses a more reusable pythonic idiom, a generator function, and is a little more compact (at the slight cost of an extra comparison):

``````def yield_non_summer(series):
in_summer = False
def stateful_summer_predicate(v):
nonlocal in_summer
if in_summer and v == 9:
in_summer = False
return True  # 9 is still in summer
elif not in_summer and v == 6:
in_summer = True
return in_summer
return (v for v in series if not stateful_summer_predicate(v))

def summer_69(series):
return sum(yield_non_summer(series))
``````

Or, in fewer lines:

``````def yield_non_summer(series):
in_summer = False
def stateful_summer_predicate(v):
nonlocal in_summer
in_summer = (in_summer or v == 6) and v != 9
return in_summer
return (v for v in series if not stateful_summer_predicate(v))

def summer_69(series):
return sum(yield_non_summer(series))
``````

Something like this:

``````def summer_69(lst):
"""Return the sum of the numbers in the array,
except ignore sections of numbers starting with a 6 and extending to the next 9
(every 6 will be followed by at least one 9). Return 0 for no numbers
"""
if not lst:
return 0
else:
_sum = 0
active = True
for x in lst:
if active:
if x != 6:
_sum += x
else:
active = False
else:
if x == 9:
active = True
return _sum

print(summer_69([1, 3, 5]))
print(summer_69([4, 5, 6, 7, 8, 9]))
print(summer_69([2, 1, 6, 9, 11]))
``````

output

``````9
9
14
``````
``````def summer_69(arr):
if 6 in arr:
c=arr[arr.index(6):arr.index(9)+1]
for i in c:
arr.remove(i)
print(arr)
return sum(arr)
else:
return sum(arr)

summer_69([1,2,3,4,5,6,7,8,9,10,11,12])
``````

This will work:

``````def summer_69(arr):
total = 0
for num in arr:
if num != 6:
total += num
break
else:
if num != 9:
break
else:
break
``````
``````def summer_69(arr):
sum = 0
Flag = False
if 6 not in arr:
for num in arr:
sum = sum + num
return sum
else:
for num in arr:
if num != 6 and Flag == False:
sum = sum + num
elif num == 6:
Flag = True
continue
elif Flag == True and num != 9:
continue
elif num == 9:
Flag = False
return sum
``````

A simple approach is to make a filter and sum the results.

Code

``````def filter_substring(seq, start, end):
"""Yield values outside a given substring."""
release = True

for x in seq:

if x == start:
release = False
elif x == end:
release = True
elif release:
yield x

def summer(seq):
"""Return the sum of certain values."""
return sum(filter_substring(seq, 6, 9))
``````

Demo

``````assert 0 == summer([])
assert 6 == summer([1, 2, 3])
assert 6 == summer([1, 2, 3, 6, 8, 7, 9])
assert 9 == summer([1, 3, 5])
assert 8 == summer([3, 5, 6, 7, 8, 9])
assert 15 == summer([2, 1, 6, 9, 12])
assert 16 == summer([2, 1, 6, 9, 1, 6, 6, 120, 9, 9, 12])

``````

Details

`filter_substring()`+

This is a generator function. It iterates the input sequence and only yields a value if the conditions are appropriate, i.e. when the `release` remains `True`.

``````>>> list(filter_substring("abcde", "c", "d"))
['a', 'b', 'e']
>>> list(filter_substring([0, 1, 2, 3, 10], 1, 3))
[0, 10]
``````

`summer()`

Here we simply sum whatever `filter_range()` yields.

+Note: a substring is a contiguous subsequence; this may or may not include strings in Python.

``````def summer_69(arr):
a = 0
for nums in arr:
if nums == 6:
for items in arr[arr.index(6):]:
a = a+ items
if items == 9:
break
return sum(arr)-a
``````
``````def summer_69(arr):
x = arr.count(6)
y = arr.count(9)
# to decide number of iteration required for loop
z = min(x,y)
k = 0
while k < (z) :
m = arr.index(6)
n = arr.index(9)
del arr[m:(n+1)]
k = k + 1
print(arr)
return sum(arr)
``````

This will work for summer_69 problem as well for filtering substring

``````def filter_substring(seq, start, end):
flag = False
for char in seq:
if char == start:
flag = True
continue
elif flag:
if char == end:
flag = False
else:
continue
else:
yield char

def summer_69(seq, start, end):
return sum(filter_substring(seq, start, end))

def print_substring(string, start, end):
return list(filter_substring(string, start, end))
``````

## Example ::

``````seq = [4, 5, 9, 6, 2, 9, 5, 6, 1, 9, 2]
print(summer_69(seq, start=6, end=9))

string = "abcdef"
print(print_substring(string, start='c', end='e'))
``````

Will work with indexes:

``````def summer_69(arr):
y = []
for x in arr:
if 6 in arr:
a = arr.index(6)
b = arr.index(9)
del arr[a:b+1]
y = arr
elif arr == []:
return "0"
else:
return sum(arr)
return sum(y)
print(summer_69([]))                                                          #0
print(summer_69([1, 3, 5]))                                                   #9
print(summer_69([4, 5, 6, 7, 8, 9]))                                          #9
print(summer_69([2, 1, 6, 9, 11]))                                            #14
print(summer_69([2, 1, 6, 9, 6, 11, 25, 36, 11, 9, 4, 6, 4, 6, 3, 9, 15]))    #22
``````

This is the probably best answer if you are a newbie. I have simplified it as much as i can. you only need to know enumerate, function, for loops , tuple unpacking,if/else statements and break function.So lets go straight to the answer.

``````def myfunc(a):
mylist=[]
sum1 = 0
for b,c in enumerate(a):
if c==6:
for d in  a[:b]:
mylist.append(d)
for e,f in enumerate(a):
if f==9:
for j in a[e+1:]:
mylist.append(j)
for y in a:
if y==6:
break
else:
mylist.append(y)
for k in mylist:
sum1 = sum1+k
print(sum1)
``````

myfunc([1,3,5])

Replace

``````list1.index(9)+1
by
list1.index(9,list1.index(6)+1)+1
``````

in line 6.
This will start searching for a 9 after 6.

For those interested, here is my solution for this problem:

``````def summer_69(arr):
skate = arr
guitar = []
for i in range(len(arr)):
if 6 in arr:
guitar = skate[skate.index(6):skate.index(9)+1]
return abs(sum(skate) - sum(guitar))
else:
return sum(skate)
``````

Short O(n) solution using an iterator and the `in` operator to search for (and thereby skip to) the 9 following each 6:

``````def summer_69(lst):
it = iter(lst)
return sum(x for x in it
if x != 6 or 9 not in it)
``````

Less dense version:

``````def summer_69(lst):
it = iter(lst)
total = 0
for x in it:
if x == 6:
9 in it
else:
total += x
``````

Correctness check (random test cases) and benchmark (with `[1] * 5000 + [6, 9] * 2500`) along with the accepted answer’s solution (which takes O(n2)):

``````30 out of 30 tests correct

303045 us  303714 us  304335 us  306007 us  309986 us  summer_69_Accepted
444 us     446 us     464 us     478 us     527 us  summer_69_Kelly1
442 us     448 us     453 us     465 us     500 us  summer_69_Kelly2
``````

Code (Try it online!):

``````from timeit import repeat

def summer_69_Accepted(lst):
copyoflist = lst[:] # makes shallow copy of list
while True:
if 6 not in copyoflist:
return sum(copyoflist)

indexof6 = copyoflist.index(6)
indexof9 = copyoflist.index(9, indexof6+1) # begin search for 9 after 6
del copyoflist[indexof6:indexof9+1]

def summer_69_Kelly1(lst):
it = iter(lst)
return sum(x for x in it
if x != 6 or 9 not in it)

def summer_69_Kelly2(lst):
it = iter(lst)
total = 0
for x in it:
if x == 6:
9 in it
else:
total += x

funcs = summer_69_Accepted, summer_69_Kelly1, summer_69_Kelly2

from random import randrange, choices

def testcase():
def others():
return choices([0, 1, 2, 3, 4, 5, 7, 8], k=randrange(10))
lst = others()
for _ in range(10):
lst += [6, *others(), 9, *others()]
return lst

tests = correct = 0
for _ in range(10):
lst = testcase()
expect = funcs[0](lst.copy())
for func in funcs:
result = func(lst.copy())
correct += result == expect
tests += 1
print(correct, 'out of', tests, 'tests correct')
print()

lst = [1] * 5000 + [6, 9] * 2500
for func in funcs:
times = repeat(lambda: func(lst), number=1)
print(*('%6d us ' % (t * 1e6) for t in sorted(times)), func.__name__)
``````

This is taken from a Udemy course.

Here is the official answer . . .

``````def summer_69(arr):
total = 0
for num in arr:
if num != 6:
total += num
break
else:
if num != 9:
break
else:
break
``````

Jose Periera has this on Python ‘zero to hero’ course.

``````def summer_69(arr):

#first find out if 6 or 9 are in the list

if 6 in arr and 9 in arr:

#Then create a variable that stores the index location of the number 6
#and the number 9

sixer = arr.index(6)
niner = arr.index(9)

#now return the sum of the array minus the sum of the values between
#index of 6 and index of 9 inclusive (hence the plus 1)
#This way will ignore the case of a 9 appearring before a 6 too.

return sum(arr) - sum(arr[sixer:niner+1])

#Otherwise just return the sum of the array.

else:
return sum(arr)
``````

Happy to accept criticism here. I’m learning Python myself and I’m undergoing an Msc in computer science and hoping to apply for jobs in the field soon, so your comments will help me 🙂

My approach was to sum the whole list and the part of the list that we want to ignore and subtract them at the end.

``````def summer_69(arr):
result=0
reduction =0
for i in range(0,len(arr)):
result+=arr[i]
if arr[i] == 6:
temp = arr[arr.index(6):arr.index(9)+1]
reduction = sum(temp)
return result - reduction
``````
``````def summer69(a):
for z in a:
if z==6 and 9 in a:
x=a.index(6)
y=a.index(9)
del a[x:y+1]
t= sum(a)
else:
t=sum(a)
return t
``````

Will always prefer short , clear and easy understandable solution.

``````def summer_69(arr):
if 9 in arr :
sum = 0
y = arr.index(9)
for i , l in enumerate(arr):
if l == 6:
del arr[i:y+1]
for i in range(len(arr)):
sum = sum + arr[i]
return sum
elif 9 not in arr:
sum = 0
for i in range(len(arr)):
sum = sum + arr[i]
return sum
``````

def summer_69(mylist):

``````if 6 in mylist:
return sum(mylist) - sum(mylist[mylist.index(6):mylist.index(9)+1])
else:
return sum(mylist)
``````

def summer_69(arr):

``````returner = []

if 6 and 9 in arr:

a = arr.index(6)
b = arr.index(9)

if a < b:

seq = arr[a:(b+1)]

for i in arr:
if i not in seq:
returner.append(i)
return (sum(returner))

elif a > b:

seq = arr[b:(a+1)]

for i in arr:
if i not in seq:
returner.append(i)
return (sum(returner))

elif 6 in arr:

a = arr.index(6)
seq = arr[a:]
for i in arr:
if i not in slicer:
returner.append(i)
return(sum(returner))

elif 9 in arr:
a = arr.index(9)
seq = arr[a:]
for i in arr:
if i not in slicer:
returner.append(i)
return(sum(returner))

elif arr == []:
return 0

else:
return (sum(arr))
``````

Just learning Python too and this was what I came up with for it:

``````def myfunc(arr):
ignore_list = []
newlist = []
for i,v in enumerate(arr):
if v >= 6 and v <= 9:
ignore_list.append(i)
if i in ignore_list:
newlist.append(0)
else:
newlist.append(v)

return sum(newlist)
``````
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