Styjun

I'm trying to increment through a list and replace multiple occurrences (in sequence) of a given value with the amount of times they occurred (as a rough range, with occurrences of 1-3 being small, 4-6 being medium, and more than 6 being a large amount of times).

I am trying to find an elegant solution to this problem and hoping for any guidance. I looked through itertools but couldn't find something suitable for this (I think).

Any help would be greatly appreciated. Thank you

So for instance:

testList = [1, 2, 2, 1, 1, 2, 1, 4, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1]

will become

[1, "2 small amount of times", 1, 1, "2 small amount of times", 1, 4, 1, "2 medium amount of times", 1, "2 large amount of times"]

testList = [1, 2, 2, 1, 1, 2, 1, 4, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1]

listLocation = -1

newlist = []

for i in testList:

 listLocation += 1

 if i == 2:

 if testList[listLocation+1] == 2:

 testList[listLocation] = "2 multiple"

 newlist.append(testList[listLocation])

 testList.pop(listLocation+1)

 else:
 newlist.append(i)

newlist

This is as far as I've gotten, right now this just detects when 2 occurs multiple times in sequence and replaces that sequence with a string, but I can't work out how to move from this to an actual counter that bins it by ranges and a more elegant style of code (I'm sure there is a way to avoid having the listLocation variable to keep track of list index). Also I can't work out how to detect the end of the list because right now this will crash if it hits a 2 as the last value in the list.

Any help would be greatly appreciated, thank you

Here is my solution. First, let's define a function which returns a message to be inserted into the list:

def message(value, count):
 msg = 'value amount amount of times'

 if 1 <= count <= 3:
 msg = msg.format(value=value, amount='small')
 elif 4 <= count <= 6:
 msg = msg.format(value=value, amount='medium')
 else:
 msg = msg.format(value=value, amount='large')

 return msg

Second, we define a function which takes a list of values and a value to be counted as its arguments:

def counter(values, value):
 count = 0
 results = []

 for i, v in enumerate(values):

 if v != value:

 if count:
 results.append(message(value, count))
 count = 0

 results.append(v)
 continue

 count = count + 1

 if i < len(values) - 1:
 continue

 results.append(message(value, count))

 return results

Here is the result:

>>> values = [1, 2, 2, 1, 1, 2, 1, 4, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1]
>>> counter(values, value=2)
[1,
 '2 small amount of times',
 1,
 1,
 '2 small amount of times',
 1,
 4,
 1,
 '2 medium amount of times',
 1,
 '2 large amount of times',
 1]

Here's what I think is a simple way to do it.

a = [1, 2, 2, 1, 1, 2, 1, 4, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1]
b = []
i=0
while i<len(a) and i < len(a):
 if a[i]==2:
 n=i
 while a[i]==2:
 i+=1

 if (i-n)<4:
 b.append("2 small amount of times")
 elif (i-n)<6 and (i-n)>4:
 b.append("2 medium amount of times")
 else:
 b.append("2 large amount of times") 

 else:
 b.append(a[i])
 i+=1
print(b)

Outputs:

[1, '2 small amount of times', 1, 1, '2 small amount of times', 1, 4, 1, '2 large amount of times', 1, '2 large amount of times', 1]

Let me know if you have any queries.

If you want to group all the numbers in the list, you can first create a list of lists, and then regroup them.

First to create a list of lists with the same number:

def group_list(input_list):
 l,s = [], 0
 for i in range(len(input_list)):
 try:
 if input_list[i] != input_list[i+1]:
 l.append(input_list[s:i+1])
 s=i+1
 except IndexError:
 if input_list[i] == input_list[i-1]:
 l.append(input_list[s:i+1])
 else:
 l.append([input_list[i]])
 return l

Then apply the amount message:

def message(num):
 if num > 6:
 return 'large'
 elif num >4:
 return 'medium'
 else:
 return 'small'

With both functions ready, do a list comprehension.

another_list = [i[0] if len(i)<2 else f"i[0] message(len(i)) amount of times" for i in group_list(testList)]

print (another_list)

Result:

[1, '2 small amount of times', '1 small amount of times', 2, 1, '4 small amount of times', 1, '2 medium amount of times', 1, '2 large amount of times', 1]