Fetching values with consecutive keys from dictionary

Question

I have dictionary like:

{0:'I', 3: 'Mr.', 5: 'Nevrekar', 8: 'JK', 14: 'Soham', 15: 'Sharma',  4: 'Akshay'}

I want to return it the 1st combination of consecutive keys(minimum 2). In above case it's (3,4,5).

so it would return a list:

['Mr.', 'Akshay', 'Nevrekar'].

How can I achieve this?

My Attempt:

name=[]
count = 0
for k,g in name_list.items():
    if count == 0:
        temp =None
    if k-1 == temp:
        name.append(g)

    count+= 1
    temp = k

which is giving output as:

['Akshay', 'Nevrekar', 'Sharma']

Dictionaries aren't guarranteed to be ordered. In python 3.7 - "the insertion-order preservation nature of dict objects has been declared to be an official part of the Python language spec." — Sayse, Aug 16 '18 at 06:41
I just want the 1st combination of consecutive keys. It should ignore the rest. — Sociopath, Aug 16 '18 at 06:41
After ['Mr.', 'Akshay', 'Nevrekar'] what is next in sequence? — Prayson W. Daniel, Aug 16 '18 at 06:42
@PraysonDaniel For above input that;s it. It should not return anything after that(Ideally break the loop). — Sociopath, Aug 16 '18 at 06:43
https://stackoverflow.com/questions/2361945/detecting-consecutive-integers-in-a-list — U13-Forward, Aug 16 '18 at 06:44

Ma0 · Accepted Answer · 2018-08-16T06:50:31.600

1

There is a recipe from more_itertools that comes in handy here:

import more_itertools as mit

d = {0:'I', 3: 'Mr.', 4: 'Akshay', 5: 'Nevrekar', 8: 'JK', 14: 'Soham', 15: 'Sharma'}
cons = [list(group) for group in mit.consecutive_groups(sorted(d.keys()))]
# [[0], [3, 4, 5], [8], [14, 15]]
res = [[d[y] for y in x] for x in cons if len(x)>1]
# [['Mr.', 'Akshay', 'Nevrekar'], ['Soham', 'Sharma']]

If you only need groups where the consecutive integers are at least 3 you have to modify the inequality accordingly (len(x) > 2).

edited Aug 16 '18 at 06:50

answered Aug 16 '18 at 06:43

Ma0

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it's not working when dictionary is not ordered. For input `name_list = {0:'I', 3: 'Mr.', 5: 'Nevrekar', 8: 'JK', 14: 'Soham', 15: 'Sharma', 4: 'Akshay'}` – Sociopath Aug 16 '18 at 06:48
Thanks :-) This is what I was looking for. – Sociopath Aug 16 '18 at 06:54

score 0 · Answer 2 · answered Aug 16 '18 at 06:58

0

From the docs:

>>> a = {0:'I', 3: 'Mr.', 4: 'Akshay', 5: 'Nevrekar', 8: 'JK', 14: 'Soham', 15: 'Sharma'}
>>>
>>> 
>>> from itertools import groupby
>>> from operator import itemgetter
>>>
>>> con_ser = None
>>> for k, g in groupby(enumerate(sorted(a.keys())), lambda (i,x):i-x):
...     item = map(itemgetter(1), g)
...     if(len(item) >= 3):
...             con_ser = item
...             break
... 
>>> 
>>> [a.get(x) for x in con_ser] .        #con_ser = [3, 4, 5]
['Mr.', 'Akshay', 'Nevrekar']
>>>

answered Aug 16 '18 at 06:58

Pankaj Singhal

15,283
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getting invalid syntax at `lambda (i,x):i-x` – Sociopath Aug 16 '18 at 07:04
I've copied the code from my terminal. Should work for you. I'm on python 2.7 – Pankaj Singhal Aug 16 '18 at 07:05

Fetching values with consecutive keys from dictionary

2 Answers2