How to split string array to 2-dimension char array in python - python

I have a string array, for example:
a = ['123', '456', '789']
I want to split it to form a 2-dimension char array:
b = [['1', '2', '3'], ['4', '5', '6'], ['7', '8', '9']]
I'm using
[[element for element in line] for line in array]
to achieve my goal but found it not easy to read, is there any built-in function or any readable way to do it?


Looks like a job for map:
>>> a = ['123', '456', '789']
>>> map(list, a)
[['1', '2', '3'], ['4', '5', '6'], ['7', '8', '9']]
Relevant documentation:
map
list


you could do something like:
first_list = ['123', '456', '789']
other_weirder_list = [list(line) for line in first_list]
Your solution isn't that bad, but you might do something like this or the map suggestion by arashajii.


map(list, array) should do it.


You can use map:
>>> a
['123', '456', '789']
>>> map(list, a)
[['1', '2', '3'], ['4', '5', '6'], ['7', '8', '9']]
Although I really don't see why you'd need to do this (unless you plan on editing one specific character in the string?). Strings behave similarly to lists.


First I tried e.split(''), but I get ValueError: empty separator.
Try this:
a = ['123', '456', '789']
b = [list(e) for e in a]
b
[['1', '2', '3'], ['4', '5', '6'], ['7', '8', '9']]

Related

Remove sublist duplicates including reversed

For example i have following
list = [['1', '2'], ['1', '3'], ['1', '4'], ['1', '5'], ['2', '1'], ['4', '1'], ['2', '6']]
I want to match if a sub list has a reversed sub list within same list (i.e. ['1', '2'] = ['2', '1']) , and if True than to remove from the list the mirrored one.
The final list should look like :
list = [['1', '2'], ['1', '3'], ['1', '4'], ['1', '5']['2', '6']]
This is what i tried:
for i in range(len(list)):
if list[i] == list[i][::-1]:
print("Match found")
del list[i][::-1]
print(list)
But finally I get the same list as original. I am not sure if my matching condition is correct.

You could iterate over the elements of the list, and use a set to keep track of those that have been seen so far. Using a set is a more convenient way to check for membership, since the operation has a lower complexity, and in that case you'll need to work with tuples, since lists aren't hashable. Then just keep those items if neither the actual tuple or the reversed have been seen (if you just want to ignore those which have a reversed you just need if tuple(reversed(t)) in s):
s = set()
out = []
for i in l:
t = tuple(i)
if t in s or tuple(reversed(t)) in s:
continue
s.add(t)
out.append(i)
print(out)
# [['1', '2'], ['1', '3'], ['1', '4'], ['1', '5'], ['2', '6']]

lists = [['1', '2'], ['1', '3'], ['1', '4'], ['1', '5'], ['2', '1'], ['4', '1'], ['2', '6']]
for x in lists:
z=x[::-1]
if z in lists:
lists.remove(z)
Explanation: While looping over lists, reverse each element and store in 'z'. Now, if 'z' exists in lists, remove it using remove()
The problem with your solution is you are checking while using index 'i' which means if an element at 'i' is equal to its reverse which can never happen!! hence getting the same results

Approach1:
new_list = []
for l in List:
if l not in new_list and sorted(l) not in new_list:
new_list.append(l)
print(new_list)
Approach2:
You can try like this also:
seen = set()
print([x for x in List if frozenset(x) not in seen and not seen.add(frozenset(x))])
[['1', '2'], ['1', '3'], ['1', '4'], ['1', '5'], ['2', '6']]

my_list = [['1', '2'], ['1', '3'], ['1', '4'], ['1', '5'], ['2', '1'], ['4', '1'], ['2', '6']]
my_list = list(set([sorted(l) for l in my_list]))

This is similar to solution by #Mehul Gupta, but I think their solution is traversing the list twice if matched: one for checking and one for removing. Instead, we could
the_list = [['1', '2'], ['1', '3'], ['1', '4'], ['1', '5'], ['2', '1'], ['4', '1'], ['2', '6']]
for sub_list in the_list:
try:
idx = the_list.index(sub_list[::-1])
except ValueError:
continue
else:
the_list.pop(idx)
print(the_list)
# [['1', '2'], ['1', '3'], ['1', '4'], ['1', '5'], ['2', '6']]
because it is easier to ask for forgiveness than permission.
Note: Removing elements whilst looping is not a good thing but for this specific problem, it does no harm. In fact, it is better because we do not check the mirrored again; we already removed it.

As I have written in a comment, do never use list (or any built-in) as a variable name:
L = [['1', '2'], ['1', '3'], ['1', '4'], ['1', '5'], ['2', '1'], ['4', '1'], ['2', '6']]
Have a look at your code:
for i in range(len(L)):
if L[i] == L[i][::-1]:
print("Match found")
del L[i][::-1]
There are two issues. First, you compare L[i] with L[i][::-1], but you want to compare L[i] with L[j][::-1] for any j != i. Second, you try to delete elements of a list during an iteration. If you delete an element, then the list length is decreased and the index of the loop will be out of the bounds of list:
>>> L = [1,2,3]
>>> for i in range(len(L)):
... del L[i]
...
Traceback (most recent call last):
...
IndexError: list assignment index out of range
To fix the first issue, you can iterate twice over the elements: for each element, is there another element that is the reverse of the first? To fix the second issue, you have two options: 1. build a new list; 2. proceed in reverse order, to delete first the last indices.
First version:
new_L = []
for i in range(len(L)):
for j in range(i+1, len(L)):
if L[i] == L[j][::-1]:
print("Match found")
break
else: # no break
new_L.append(L[i])
print(new_L)
Second version:
for i in range(len(L)-1, -1, -1):
for j in range(0, i):
if L[i] == L[j][::-1]:
print("Match found")
del L[i]
print(L)
(For a better time complexity, see #yatu's answer.)
For a one-liner, you can use the functools module:
>>> L = [['1', '2'], ['1', '3'], ['1', '4'], ['1', '5'], ['2', '1'], ['4', '1'], ['2', '6']]
>>> import functools
>>> functools.reduce(lambda acc, x: acc if x[::-1] in acc else acc + [x], L, [])
[['1', '2'], ['1', '3'], ['1', '4'], ['1', '5'], ['2', '6']]
The logic is the same as the logic of the first version.

You can try this also:-
l = [['1', '2'], ['1', '3'], ['1', '4'], ['1', '5'], ['2', '1'], ['4', '1'], ['2', '6']]
res = []
for sub_list in l:
if sub_list[::-1] not in res:
res.append(sub_list)
print(res)
Output:-
[['1', '2'], ['1', '3'], ['1', '4'], ['1', '5'], ['2', '6']]

Convert List of string to Integer

I have a list whose elements are only single string.
my_list = ['9', '4', '4', '5', '4', '3', '5']
How can I convert this into a single integer like: 9445435
Note: ' '.join() only works for lists of strings, not integers.

Try this,
In [5]: int(''.join(my_list))
Out[5]: 9445435

Why don't you try this approach?
my_list = ['9', '4', '4', '5', '4', '3', '5']
output = int(''.join(str(num) for num in my_list))
print(output) ## 9445435

result = ' '.join(my_list)
result = int(result)

How to create lists in list where every future list is separate by spaces in list

User provides input with spaces:
row = list(input())
print(row)
['1','2','3',' ','4','5','6',' ','7','8','9',' ']
So I need to create 'row' list into the below. The list is divided into sub-lists based on whitespace:
[['1','2','3'],['4','5','6'],['7','8','9']]

You can use str.split to split by whitespace:
myinput = '123 456 789'
row = list(map(list, myinput.split()))
print(row)
[['1', '2', '3'], ['4', '5', '6'], ['7', '8', '9']]
Alternatively, using a list comprehension:
row = [list(i) for i in myinput.split()]

You can usestr.split to split the input on spaces to give a list of sub-strings.
E.g. '123 456 789' would become ['123', '456', '789'].
Then use a list-comprehension to convert these strings into lists of characters with the list() constructor (as you are already familiar with).
Making the final code:
row = [list(s) for s in input().split()]
#[['1', '2', '3'], ['4', '5', '6'], ['7', '8', '9']]

Starting with your list rather than the string, you can do that using itetools.groupby:
from itertools import groupby
row = ['1','2','3',' ','4','5','6',' ','7','8','9',' ']
out = [list(group) for key, group in groupby(row, lambda x: x != ' ') if key]
print(out)
# [['1', '2', '3'], ['4', '5', '6'], ['7', '8', '9']]
We group the values depending on whether or not they are spaces, and only keep the groups that are not made of spaces.

Try this:
abc=['1','2','3',' ','4','5','6',' ','7','8','9',' ']
newList=list()
temp=list()
for i in abc:
if(i==' '):
newList.append(temp)
temp=list()
else:
temp.append(i)
print(newList)
Output:
[['1', '2', '3'], ['4', '5', '6'], ['7', '8', '9']]

Remove duplicate items from list

I tried following this post but, it doesnt seem to be working for me.
I tried this code:
for bresult in response.css(LIST_SELECTOR):
NAME_SELECTOR = 'h2 a ::attr(href)'
yield {
'name': bresult.css(NAME_SELECTOR).extract_first(),
}
b_result_list.append(bresult.css(NAME_SELECTOR).extract_first())
#set b_result_list to SET to remove dups, then change back to LIST
set(b_result_list)
list(set(b_result_list))
for brl in b_result_list:
print("brl: {}".format(brl))
This prints out:
brl: https://facebook.site.com/users/login
brl: https://facebook.site.com/users
brl: https://facebook.site.com/users/login
When I just need:
brl: https://facebook.site.com/users/login
brl: https://facebook.site.com/users
What am I doing wrong here?
Thank you!

you are discarding the result when you need to save it ... b_result_list never actually changes... so you are just iterating over the original list. instead save the result of the set operation
b_result_list = list(set(b_result_list))
(note that sets do not preserve order)

If you want to maintain order and uniqueify, you can do:
>>> li
['1', '1', '2', '2', '3', '3', '3', '3', '1', '1', '4', '5', '4', '6', '6']
>>> seen=set()
>>> [e for e in li if not (e in seen or seen.add(e))]
['1', '2', '3', '4', '5', '6']
Or, you can use the keys of an OrderedDict:
>>> from collections import OrderedDict
>>> OrderedDict([(k, None) for k in li]).keys()
['1', '2', '3', '4', '5', '6']
But a set alone may substantially change the order of the original list:
>>> list(set(li))
['1', '3', '2', '5', '4', '6']

How could i refresh a list once an item has been removed from a list within a list in python

This is quite complicated but i would like to be able to refresh a larger list once at item has been taken out of a mini list within the bigger list.
listA = ['1','2','3','4','5','6','6','8','9','5','3','7']
i used the code below to split it into lists of threes
split = [listA[i:(i+3)] for i in range(0, len(listA) - 1, 3)]
print(split)
# [['1','2','3'],['4','5','6'],['6','8','9'],['5','3','7']]
split = [['1','2','3'],['4','5','6'],['6','8','9'],['5','3','7']]
if i deleted #3 from the first list, split will now be
del split[0][-1]
split = [['1','2'],['4','5','6'],['6','8','9'],['5','3','7']]
after #3 has been deleted, i would like to be able to refresh the list so that it looks like;
split = [['1','2','4'],['5','6','6'],['8','9','5'],['3','7']]
thanks in advance

Not sure how big this list is getting, but you would need to flatten it and recalculate it:
>>> listA = ['1','2','3','4','5','6','6','8','9','5','3','7']
>>> split = [listA[i:(i+3)] for i in range(0, len(listA) - 1, 3)]
>>> split
[['1', '2', '3'], ['4', '5', '6'], ['6', '8', '9'], ['5', '3', '7']]
>>> del split[0][-1]
>>> split
[['1', '2'], ['4', '5', '6'], ['6', '8', '9'], ['5', '3', '7']]
>>> listA = sum(split, []) # <- flatten split list back to 1 level
>>> listA
['1', '2', '4', '5', '6', '6', '8', '9', '5', '3', '7']
>>> split = [listA[i:(i+3)] for i in range(0, len(listA) - 1, 3)]
>>> split
[['1', '2', '4'], ['5', '6', '6'], ['8', '9', '5'], ['3', '7']]

Just recreate the single list from your nested lists, then re-split.
You can join the lists, assuming they are only one level deep, with something like:
rejoined = [element for sublist in split for element in sublist]
There are no doubt fancier ways, or single-liners that use itertools or some other library, but don't overthink it. If you're only talking about a few hundred or even a few thousand items this solution is quite good enough.

I need this for turning of cards in the deck in a solitaire game.
You can deal your cards using itertools.groupby() with a good key function:
def group_key(x, n=3, flag=[0], counter=itertools.count(0)):
if next(counter) % n == 0:
flag[0] = flag[0] ^ 1
return flag[0]
^ is a bitwise operator, basically it change the value of the flag from 0 to 1 and viceversa. The flag value is an element of a list because we're doing some kind of memoization.
Example:
>>> deck = ['1', '2', '3', '4', '5', '6', '6', '8', '9', '5', '3', '7']
>>> for k,g in itertools.groupby(deck, key=group_key):
... print(list(g))
['1', '2', '3']
['4', '5', '6']
['6', '8', '9']
['5', '3', '7']
Now let's say you've used card '9' and '8', so your new deck looks like:
>>> deck = ['1', '2', '3', '4', '5', '6', '6', '5', '3', '7']
>>> for k,g in itertools.groupby(deck, key=group_key):
... print(list(g))
['1', '2', '3']
['4', '5', '6']
['6', '5', '3']
['7']

Build an object that contains a list and tracks when the list is altered (probably by controlling write to it), then have the object do it's own split every time the data is altered and save the split list to a member of the object.

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