counter in loop /list/ count same vale if they are one after another (python)

Question

The idea of the program is to count life time of some structures (count same structures which are one after another) but i have problem with implementing counter or enumerate function, I don't want to count number of all. I've tried do this many ways. Now I am stuck.

struct=[1, 2, 2, 2, 3, 3, 1, 2, 3, 1, 1, 1, 1, 2, 1, 1,] #input list of structures
list1=[] #list of lifetimes for struct 1

for i in range (0, len(struct)-1):
    a=struct[i-1]
    b=struct[i]
    c=struct[i+1]
    if a!=1 and b==1 and c!=1: # if it is one separated record
        list1.append(1)
    if a!=1 and b==1 and c==1: # for first record in sequence 
        list1.append('f')
    if a==1 and b==1 and c==1: # for middle record in sequence
        list1.append('m')
    if a==1 and b==1 and c!=1: # for last record in sequence 
        list1.append('l')

print (list1)

it gives me: [1, 1, 'f', 'm', 'm', 'l', 'f', 'l']
could you give me any advise how to implement counter ? for example ['f', 'm', 'm', 'l] (firs/middle/middle/last) is given from [1, 1, 1, 1] from the list of structure so it's 4 records to get [1, 1, 4, 2]

Sorry, for my non programming language, I am beginner in this field. I've search any similar questions but couldn't find.

Should all lines after for i in range (0, len(struct)-1): and before print (list1) be indented? — Daniel Robinson
– Daniel Robinson, Commented Jan 20, 2015 at 23:30
Why are you appending letters to your list if you want only numbers in the output? — dg99
– dg99, Commented Jan 20, 2015 at 23:33
Why are you appending, 'f', 'm', and 'l'? How are you expecting appending just 1, 'f', 'm', and 'l' to a list to give you a list that has a 2 and a 4 in it? And what is the meaning of [1, 1, 4, 2] with respect to this list anyway? — Daniel Robinson
– Daniel Robinson, Commented Jan 20, 2015 at 23:33
bad copy paste, they are all in for loop, if i clearly understand you — Asia Lach
– Asia Lach, Commented Jan 20, 2015 at 23:34

jgritty · Accepted Answer · 2015-01-20 23:57:34Z

2

Is it something like this, you're looking for?

inlist = [1, 2, 2, 2, 3, 3, 1, 2, 3, 1, 1, 1, 1, 2, 1, 1,]
#         1                 1        1--2--3--4     1--2
def lifetime(num):
    output = []
    count = 0
    for x in inlist:
        if x == num:
            count += 1
        else:
            if count != 0:
                output.append(count)
                count = 0
    if count != 0:
        output.append(count)

    print output

lifetime(1)
lifetime(2)
lifetime(3)

Output:

[1, 1, 4, 2]
[3, 1, 1]
[2, 1]

edited Jan 20, 2015 at 23:57

answered Jan 20, 2015 at 23:40

jgritty

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2 Comments

Asia Lach Over a year ago

as @augurar said I need list of group lengths for each group of consecutive 1s in the list

Asia Lach Over a year ago

YES! Thank you! Sorry for bothering in such a trivial case!

augurar · Accepted Answer · 2015-01-21 04:38:33Z

0

You can use itertools.groupby() to group the list by value, then return the counts for each group with a given value:

from itertools import groupby

def ilen(iterable):
    """ Return the length of an iterable """
    return sum(1 for _ in iterable)

def lifetime(num, lst):
    """ Return a list of counts for each group of num in lst """
    return [ilen(g) for k, g in groupby(lst) if k==1]

Example:

>>> inlist = [1, 2, 2, 2, 3, 3, 1, 2, 3, 1, 1, 1, 1, 2, 1, 1,]
>>> lifetime(1, inlist)
[1, 1, 4, 2]

answered Jan 21, 2015 at 4:38

augurar

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