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Remove an element from a lists of lists in Python [duplicate]

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How to remove items from a list while iterating?
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How can I remove from a list of list every list in which a '5' is placed before a '3' given an initial list of list like the following one?
[('3', '3', '3'), ('3', '3', '5'), ('3', '5', '3'), ('3', '5', '5'), ('5', '3', '3'), ('5', '3', '5'), ('5', '5', '3'), ('5', '5', '5')]
I tried with
for i in list_ck:
for j in range(0,2):
if (i[j]=='5' and i[j+1]=='3'):
list_ck.remove(i)
but it doesn't work

It's easier to test whether a 5 occurs before a 3 if they are strings, rather than tuples; so we can use ''.join to convert them to strings.
>>> data = [('3', '3', '3'), ('3', '3', '5'), ('3', '5', '3'), ('3', '5', '5'), ('5', '3', '3'), ('5', '3', '5'), ('5', '5', '3'), ('5', '5', '5')]
>>> [r for r in data if '53' not in ''.join(r)]
[('3', '3', '3'), ('3', '3', '5'), ('3', '5', '5'), ('5', '5', '5')]
This assumes you only want to test for a 5 immediately before a 3, and won't work for more general cases where the strings in the tuple could e.g. be '53' themselves. But it's sufficient for your example.
A more general solution is to use a regex, and join on a character like , which none of the strings will contain:
>>> data = [('5', '3', '1'), ('53', '1', '1'), ('5', '1', '3')]
>>> import re
>>> pattern = re.compile('(^|,)5,(.*,)*3(,|$)')
>>> [r for r in data if not pattern.search(','.join(r))]
[('53', '1', '1')]
Here the pattern (^|,)5,(.+,)*3(,|$) matches a 5 either at the start or after a ,, followed by a comma, followed by any number of things ending with commas, followed by a 3 which is either before a comma or the end of the string.

You could use a conditional list comprehension:
# List of Tuples (lot).
list_of_tups = [('3', '3', '3'), ('3', '3', '5'), ('3', '5', '3'), ('3', '5', '5'), ('5', '3', '3'), ('5', '3', '5'), ('5', '5', '3'), ('5', '5', '5')]
>>> [tup for tup in list_of_tups
if not any((a == '5' and b == '3') for a, b in zip(tup, tup[1:]))]
[('3', '3', '3'), ('3', '3', '5'), ('3', '5', '5'), ('5', '5', '5')]
To modify the list inplace rather than create a new list, create an index of items that need to be removed and then pop them off in reverse order.
idx = [n for n, tup in enumerate(list_of_tups)
if any((a == '5' and b == '3') for a, b in zip(tup, tup[1:]))]
for i in reversed(idx):
list_of_tups.pop(i)
>>> list_of_tups
[('3', '3', '3'), ('3', '3', '5'), ('3', '5', '5'), ('5', '5', '5')]

Try to create another list with the same arguments (dont copy them, create new) and remove() from the second list
tab = [('3', '3', '3'), ('3', '3', '5'), ('3', '5', '3'), ('3', '5', '5'), ('5', '3', '3'), ('5', '3', '5'), ('5', '5', '3'), ('5', '5', '5')]
tab2 = [('3', '3', '3'), ('3', '3', '5'), ('3', '5', '3'), ('3', '5', '5'), ('5', '3', '3'), ('5', '3', '5'), ('5', '5', '3'), ('5', '5', '5')]
for i in tab:
for j in range(0,2):
if i[j]=='5' and i[j+1]=='3':
tab2.remove(i)
break

How do I change a list in python

List: [['1', '2', '4'],['1', '4', '8'],['03', '8', '6', '1', '62', '7'],['53', '8', '2', '82']]
below code on list :
neighbor1 = [list[i:i + 2] for i in range(0, len(list), 1)]
output:
[[['1', '2', '4'],['1', '4', '8']],[['03', '8', '6', '1', '62', '7'],['53', '8', '2', '82']]]
[[['1', '2', '4'],['1', '4', '8']],[['03', '8', '6', '1', '62', '7'],['53', '8', '2', '82']]]
but i want :
[[['1','2'],['2','4']],[['1','4'],['4','8']],[['03','8'],['8','6'],['6','1'],['1','62'],['62','7']],[['53','8'],['8','2'],['2','82']]]

You were almost there, you just need to go one level deeper:
[[x[i:i+2] for i in range(len(x)-1)] for x in List]
btw never use keyword list as variable name, or you can run into some really weird things...

You need to group elements by pairs.
A classic way for that is:
for p, n in zip(your_list[:-1], your_list[1:]):
pair = p, n
Where p represents each previous element and n represents each next.
With that in hand, you’ll manage to solve your problem.
For instance:
rows = [['1', '2', '4'],
['1', '4', '8'],
['03', '8', '6', '1', '62', '7'],
['53', '8', '2', '82']]
result = [list(zip(row[:-1], row[1:]))
for row in rows]
print(result)
You get:
[[('1', '2'), ('2', '4')], [('1', '4'), ('4', '8')], [('03', '8'), ('8', '6'), ('6', '1'), ('1', '62'), ('62', '7')], [('53', '8'), ('8', '2'), ('2', '82')]]

How to get all combination from multiple lists?

I am not sure that my question is correct but I don't know how to explain it otherwords.
So I've got some lists like
a = ['11', '12']
b = ['21', '22']
c = ['31', '32']
And i need to get something like:
result = [
['11', '21', '31'],
['11', '21', '32'],
['11', '22', '31'],
['11', '22', '32'],
['12', '21', '31'],
['12', '21', '32'],
['12', '22', '31'],
['12', '22', '32']
]

You need itertools.product
which returns cartesian product of input iterables.
>>> a = ['11', '12']
>>> b = ['21', '22']
>>> c = ['31', '32']
>>>
>>> from itertools import product
>>>
>>> list(product(a,b,c))
[('11', '21', '31'), ('11', '21', '32'), ('11', '22', '31'), ('11', '22', '32'), ('12', '21', '31'), ('12', '21', '32'), ('12', '22', '31'), ('12', '22', '32')]
And you can use a list comprehension to convert tuples to lists:
>>> [list(i) for i in product(a,b,c)]
[['11', '21', '31'], ['11', '21', '32'], ['11', '22', '31'], ['11', '22', '32'], ['12', '21', '31'], ['12', '21', '32'], ['12', '22', '31'], ['12', '22', '32']]

User itertools, combinations:
import itertools
a = ['11', '12']
b = ['21', '22']
c = ['31', '32']
list(itertools.combinations(itertools.chain(a,b,c), 3))
[('11', '12', '21'), ('11', '12', '22'), ('11', '12', '31'), ('11', '12', '32'), ('11', '21', '22'), ('11', '21', '31'), ('11', '21', '32'), ('11', '22', '31'), ('11', '22', '32'), ('11', '31', '32'), ('12', '21', '22'), ('12', '21', '31'), ('12', '21', '32'), ('12', '22', '31'), ('12', '22', '32'), ('12', '31', '32'), ('21', '22', '31'), ('21', '22', '32'), ('21', '31', '32'), ('22', '31', '32')]
or product:
list(itertools.product(a,b,c))
[('11', '21', '31'), ('11', '21', '32'), ('11', '22', '31'), ('11', '22', '32'), ('12', '21', '31'), ('12', '21', '32'), ('12', '22', '31'), ('12', '22', '32')]

import itertools
list(itertools.product(a,b,c))
Or using numpy
import numpy
[list(x) for x in numpy.array(numpy.meshgrid(a,b,c)).T.reshape(-1,len(a))]

import numpy as np
np.array(np.meshgrid(a, b, c)).T.reshape(-1,3)
edit
import numpy as np
len = 3 #combination array length
np.array(np.meshgrid(a, b, c)).T.reshape(-1,len)

A solution without using any library:
def subproblem(list_a, list_b):
c = list()
for i, a_i in enumerate(list_a):
for j, b_j in enumerate(list_b):
c.append(a_i + [b_j])
return c
def multi_lists_find_all_combination(lists):
com = [[]]
if len(lists) == 1:
return lists[0]
for i in range(0, len(lists)):
print(i)
com = subproblem(com, lists[i])
return com
a = ['11', '12']
b = ['21', '22']
c = ['31', '32']
ans = multi_lists_find_all_combination([a,b,c])

How can I generate a list like this with python [closed]

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Closed 8 years ago.
Improve this question
Because of unclear description, I reformatted my question as the way #user2346536 told me, thanks #user2346536.
Questions:
I would like to get a list which contains all kinds of possibilities of elements like below:
Given a set of numbers, I want to generate all the 3-Combinations of
them. Also, In such combination any number can be replaced by its
opposite. Hence for (2,3,4) --> [3,2,4], [2,3,4] and [-2,3,4] are all
valid 'combinations' whereas [-2,2,4] or [3,3,4] are not.
Has been done perfectly by #Kasra, thanks for kind help.
....Need to learn how to give a question...

You can use itertools.product and permutations and zip :
>>> from itertools import product ,permutations
>>> p=list(permutations(['2','3','4']))
>>> l=[('-'+i,'-'+j,'-'+k) for i,j,k in permutations(['2','3','4'])]
>>> new=zip(l,p)
>>> new
[(('-2', '-3', '-4'), ('2', '3', '4')), (('-2', '-4', '-3'), ('2', '4', '3')), (('-3', '-2', '-4'), ('3', '2', '4')), (('-3', '-4', '-2'), ('3', '4', '2')), (('-4', '-2', '-3'), ('4', '2', '3')), (('-4', '-3', '-2'), ('4', '3', '2'))]
>>> list(list(product(*zip(j,i))) for i,j in new)
[[('2', '3', '4'), ('2', '3', '-4'), ('2', '-3', '4'), ('2', '-3', '-4'), ('-2', '3', '4'), ('-2', '3', '-4'), ('-2', '-3', '4'), ('-2', '-3', '-4')], [('2', '4', '3'), ('2', '4', '-3'), ('2', '-4', '3'), ('2', '-4', '-3'), ('-2', '4', '3'), ('-2', '4', '-3'), ('-2', '-4', '3'), ('-2', '-4', '-3')], [('3', '2', '4'), ('3', '2', '-4'), ('3', '-2', '4'), ('3', '-2', '-4'), ('-3', '2', '4'), ('-3', '2', '-4'), ('-3', '-2', '4'), ('-3', '-2', '-4')], [('3', '4', '2'), ('3', '4', '-2'), ('3', '-4', '2'), ('3', '-4', '-2'), ('-3', '4', '2'), ('-3', '4', '-2'), ('-3', '-4', '2'), ('-3', '-4', '-2')], [('4', '2', '3'), ('4', '2', '-3'), ('4', '-2', '3'), ('4', '-2', '-3'), ('-4', '2', '3'), ('-4', '2', '-3'), ('-4', '-2', '3'), ('-4', '-2', '-3')], [('4', '3', '2'), ('4', '3', '-2'), ('4', '-3', '2'), ('4', '-3', '-2'), ('-4', '3', '2'), ('-4', '3', '-2'), ('-4', '-3', '2'), ('-4', '-3', '-2')]]
Demo :
first we must create the permutations of your number list to have all the states :
>>> p=list(permutations(['2','3','4']))
[('2', '3', '4'), ('2', '4', '3'), ('3', '2', '4'), ('3', '4', '2'), ('4', '2', '3'), ('4', '3', '2')]
then create a permutations like above for negative numbers :
>>>[('-'+i,'-'+j,'-'+k) for i,j,k in permutations(['2','3','4'])]
[('-2', '-3', '-4'), ('-2', '-4', '-3'), ('-3', '-2', '-4'), ('-3', '-4', '-2'), ('-4', '-2', '-3'), ('-4', '-3', '-2')]
and then zip 2 preceding permutations for use them in product statement :
>>> new=zip(l,p)

Python join all combinations of elements within each list

I have a list of tuples each with two elements: [('1','11'),('2','22'),('3','33'),...n]
How would I find all the combinations of each tuple with only selecting one element of the tuple at a time?
The example results:
[[1,2,3],[11,2,3],[11,2,3],[11,22,33],[11,2,33],[11,22,3],[1,22,3],[1,22,33],[1,2,33]]`
itertools.combinations gives me all combinations but does not preserve selecting only one element from each tuple.
Thanks!

Use itertools.product:
In [88]: import itertools as it
In [89]: list(it.product(('1','11'),('2','22'),('3','33')))
Out[89]:
[('1', '2', '3'),
('1', '2', '33'),
('1', '22', '3'),
('1', '22', '33'),
('11', '2', '3'),
('11', '2', '33'),
('11', '22', '3'),
('11', '22', '33')]

Did you read the itertools documentation?
>>> import itertools
>>> l = [('1','11'),('2','22'),('3','33')]
>>> list(itertools.product(*l))
[('1', '2', '3'), ('1', '2', '33'), ('1', '22', '3'), ('1', '22', '33'), ('11', '2', '3'), ('11', '2', '33'), ('11', '22', '3'), ('11', '22', '33')]