Create a list from a list of tuples

Question

I have a list like this:

mylist = [(20, 'Start', '2008-10-10', 'TBS'),...,(20, 'End', '2008-11-09', 'NG'), 
          (21, 'Start', '2008-12-10', 'TBS'),...,(21, 'End', '2008-12-15', 'G'), 
          (22, 'Start', '2009-01-10', 'TBS'),...,(22, 'End', '2009-12-10', 'B'),..]

I put '...' in the example above to say there are other items for each id like 20, 21 and 22 in the list but I don't want them. The only items that I want are the items that include 'Start' or 'End'.(Other items have different words than these two words.)

I want to create a nested list like this:

[[20, 'Start', '2008-10-10', 'End', '2008-11-09', 'NG'] ,
 [21, 'Start', '2008-12-10', 'End', '2008-12-15', 'G'], 
 [22, 'Start', '2009-01-10', 'End', '2009-12-10', 'B']]

Here is my code:

code = 0
brr = []
for row in myList:
    if row[1] == "Start":
        arr = []
        code = row[0]
        arr.append([row[0], row[1], row[2]])
        continue

    if row[0] == code and row[1] == "End":
        arr.append([row[1], row[2], row[3]])
    brr.append(arr)
for k in brr:
    print(k)

But the problem is that it creates something like this:

[[[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[21, 'Start', '2008-12-10', 'End'], ['2008-12-15', 'G']], 
 [[21, 'Start', '2008-12-10', 'End'], ['2008-12-15', 'G']],
 [[22, 'Start', '2009-01-10', 'End'], ['2009-12-10', 'B']]]

And for each items I have multiple rows in the list. I don't know why? Sorry if my question is too long.

Are start and end always the first and last element of a row? — touch my body
– touch my body, Commented May 25, 2018 at 21:03
No, myList is a list of many tuples. But always start is before end. @user4343502 — user9831231
– user9831231, Commented May 25, 2018 at 21:06
Within each tuple, do start and end occur once and only once each? — touch my body
– touch my body, Commented May 25, 2018 at 21:06
For each tuple there is only one word. I want to say if it was start then store it to a list and then by ignoring other words I should search for end for that particular code and save it in the same list as start. @user4343502 — user9831231
– user9831231, Commented May 25, 2018 at 21:10

Kind Scorpion · Accepted Answer · 2018-05-25 21:02:36Z

1

You need to use arr.extend() function

arr = []
arr.append([1,2]) # arr = [[1,2]]
arr = []
arr.extend([1,2])  # arr = [1,2]

answered May 25, 2018 at 21:02

Kind Scorpion

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1 Comment

user2357112 Over a year ago

@TemporalWolf: arr.append takes exactly one argument; arr.append(1, 2) is invalid.

Taylor D. Edmiston · Accepted Answer · 2018-05-26 00:02:51Z

You can achieve this pretty simply with itertools.groupby as well:

import itertools
from pprint import pprint

mylist = [
    (20, 'Start', '2008-10-10', 'TBS'),
    (20, 'Foo', '2008-10-10', 'TBS'),
    (20, 'End', '2008-11-09', 'NG'),

    (21, 'Start', '2008-12-10', 'TBS'),
    (21, 'End', '2008-12-15', 'G'),

    (22, 'Start', '2009-01-10', 'TBS'),
    (22, 'End', '2009-12-10', 'B'),
]

rows = (x for x in mylist if x[1] in ('Start', 'End'))
grouped = itertools.groupby(rows, key=lambda x: x[0])
output = [[k, *next(grp)[1:3], *next(grp)[1:4]] for k, grp in grouped]
pprint(output)

Output:

[[20, 'Start', '2008-10-10', 'End', '2008-11-09', 'NG'],
 [21, 'Start', '2008-12-10', 'End', '2008-12-15', 'G'],
 [22, 'Start', '2009-01-10', 'End', '2009-12-10', 'B']]

Kacperito · Accepted Answer · 2018-05-25 21:18:13Z

0

Your brr.append(arr) is always adding an array for each row, that's why there are 6 elements in the ouput. Change brr.append(arr) to:

if arr not in brr:
    brr.append(arr)

As for the format, arr.append([row[0], row[1], row[2]]) adds a list of 3 elements, instead of 3 separate elements. Use extend instead.

Your final code should look like this:

code = 0
brr = []
for row in mylist:

    if row[1] == "Start":
        arr = []
        code = row[0]
        arr.extend([row[0], row[1], row[2]])
        # continue not needed here

    if row[0] == code and row[1] == "End":
        arr.extend([row[1], row[2], row[3]])

    if arr not in brr:
        brr.append(arr)

for k in brr:
    print(k)

edited May 25, 2018 at 21:18

answered May 25, 2018 at 21:08

Kacperito

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3 Comments

user9831231 Over a year ago

Perfect! extends works here. But I still have multiple rows for each item I don't know why. @Kacper Floriański

Kacperito Over a year ago

I literally just said that... the brr.append(arr) adds arr to brr each loop iteration. You have 6 tuples you iterate over so you will 6 elements in your output.

user9831231 Over a year ago

Perfect! Thanks.

Pyd · Accepted Answer · 2018-05-29 10:41:17Z

0

please try this,

startlist=[]
endlist=[]
for item in mylist:
    if 'Start' in list(item):
        startlist.append(list(item))
    elif 'End' in list(item):
        endlist.append(list(item))
outlist=[i+j for i,j in zip(startlist,endlist)]

answered May 29, 2018 at 10:41

Pyd

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