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During communities' detection I am trying to remove duplicates nodes from lists of lists (aimed to calculate ARI).
What I have – few dozen lists inside one list with different dimensions:
lst_of_lts= [[5192, 32896, 34357, 34976, 36683, 43315], … ,[19, 92585, 94137, 98381, 99041, 100395, 101100, 109759]]
What I am running:
import itertools
Lst_of_lts.sort()
Lst_of_lts_2 = list(k for k,_ in itertools.groupby(Lst_of_lts))
Lst_of_lts_nodops= [list(i) for i in {tuple(sorted(i)) for i in Lst_of_lts_2}]
For some reason, it doesn’t remove duplicates.
The dimensions remain the same-
Any suggestions?
Also tried many options such as:
Remove duplicate items from lists in Python lists and
Remove duplicated lists in list of lists in Python
If you are removing duplicates just in the list itself, you can use set.
a = np.random.randint(0,5,(10,10)).tolist()
a
Out[128]:
[[0, 3, 0, 2, 4, 4, 0, 0, 3, 3],
[2, 4, 0, 2, 4, 2, 2, 4, 3, 1],
[3, 2, 0, 1, 2, 0, 2, 0, 2, 1],
[3, 1, 4, 1, 0, 1, 4, 4, 3, 4],
[2, 0, 1, 1, 0, 4, 1, 4, 2, 3],
[0, 0, 1, 3, 4, 3, 1, 3, 0, 1],
[1, 2, 0, 2, 1, 3, 4, 2, 2, 0],
[3, 3, 2, 2, 0, 4, 1, 1, 0, 0],
[0, 1, 3, 0, 4, 4, 2, 1, 1, 4],
[0, 1, 4, 4, 0, 1, 3, 2, 1, 1]]
[list(set(i)) for i in a]
Out[129]:
[[0, 2, 3, 4],
[0, 1, 2, 3, 4],
[0, 1, 2, 3],
[0, 1, 3, 4],
[0, 1, 2, 3, 4],
[0, 1, 3, 4],
[0, 1, 2, 3, 4],
[0, 1, 2, 3, 4],
[0, 1, 2, 3, 4],
[0, 1, 2, 3, 4]]
Or if you want to preserve the order of the element, you can use dict.fromkeys
[list(dict.fromkeys(i)) for i in a]
Out[133]:
[[0, 3, 2, 4],
[2, 4, 0, 3, 1],
[3, 2, 0, 1],
[3, 1, 4, 0],
[2, 0, 1, 4, 3],
[0, 1, 3, 4],
[1, 2, 0, 3, 4],
[3, 2, 0, 4, 1],
[0, 1, 3, 4, 2],
[0, 1, 4, 3, 2]]
My data is as follows,
data = [[2, 1, 2, 2], [2, 2, 1, 5], [1, 2, 2, 2], [2, 1, 2, 5], [2, 5, 2, 1]]
I would like to transform this such that there is a 0 at 0, 1, 2, 3 and 4th positions of the internal lists and get it look like below,
new_Data = [[0, 2, 1, 2, 2], [2, 0, 2, 1, 5], [1, 2, 0, 2, 2], [2, 1, 2, 0, 5], [2, 5, 2, 1, 0]]
I have tried using the following method,
a = 0
for n in range(len(mRco1)-1):
mRco1[n][n] = [a]
But it does not seem to work.
Can anyone suggest how can I go about this?
Use the list.insert() method
for i in range(len(data)):
data[i].insert(i, 0)
result :
print(data)
>>> [[0, 2, 1, 2, 2], [2, 0, 2, 1, 5], [1, 2, 0, 2, 2], [2, 1, 2, 0, 5], [2, 5, 2, 1, 0]]
You'd like to iterate over the lists in data, and for the n'th list, insert a 0 at position n. You can use the insert function for that, and define the following loop:
for i in range(len(data)):
data[i].insert(i, 0)
So i have nested loops and array
[[0, 1], [0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4]]:
for x in string_list:
for y in string_list:
print(x,y)
provides me the output
[0, 1] [0, 1]
[0, 1] [0, 1, 2, 3, 4, 5, 6]
[0, 1] [0, 1, 2, 3, 4]
[0, 1, 2, 3, 4, 5, 6] [0, 1]
[0, 1, 2, 3, 4, 5, 6] [0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 4, 5, 6] [0, 1, 2, 3, 4]
[0, 1, 2, 3, 4] [0, 1]
[0, 1, 2, 3, 4] [0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 4] [0, 1, 2, 3, 4]
But i have a lot of duplicates pairs and i did:
for x in range(0, len(string_list)):
for y in range(x+1, len(string_list)):
print(x,y, string_list)
but it's working only for 2 digit pairs.
So what i want is:
[0, 1] [0, 1]
[0, 1] [0, 1, 2, 3, 4, 5, 6]
[0, 1] [0, 1, 2, 3, 4]
**[0, 1, 2, 3, 4, 5, 6] [0, 1]** // avoid to output that pair cause we had that one
[0, 1, 2, 3, 4, 5, 6] [0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 4, 5, 6] [0, 1, 2, 3, 4]
[0, 1, 2, 3, 4] [0, 1]
**[0, 1, 2, 3, 4] [0, 1, 2, 3, 4, 5, 6]** // avoid to output that pair cause we had that one
[0, 1, 2, 3, 4] [0, 1, 2, 3, 4]
does it possible to do without using itertools ?
Thank you!
for k, x in enumerate(string_list):
for y in string_list[k:]:
print(x,y)
You can use itertools.combinations:
for x, y in it.combinations(string_list, 2):
# process x, y
Obviously using itertools.combinations is ideal, but since you said you don't want to use itertools, you could use a set comprehension to build the set of unique combinations (you have to convert the lists to tuples to make them hashable), then convert them back to lists as needed:
[list(list(t) for t in f) for f in {
frozenset((tuple(x), tuple(y))) for y in string_list for x in string_list
}]
You can put a continue statement in the inner loop to skip duplicates:
for x in string_list:
for y in string_list:
if x == y:
continue
print(x,y)
I wanna get values from multiple(?) list, which is not 0(I think filter can be one of solutions).
list as below:
>>ls = [[i for i in np.random.randint(0, 5, 5)] for _ in range(7)]
>>ls
>>
[[2, 3, 3, 0, 0],
[4, 2, 4, 3, 2],
[1, 2, 4, 2, 4],
[2, 3, 4, 3, 1],
[0, 1, 0, 3, 0],
[3, 4, 4, 4, 3],
[3, 4, 3, 3, 2]]
Expected result is:
[2,3,3,4,2,4,3,2,1,2,4,2,4,2,3,4,3,1,1,3,3,4,4,4,3,3,4,3,3,2]
I tried using filter function, wanted expand this idea, but I failed:
>> [elem for elem in filter(lambda x: x if x != 0 else False, ls[0])]
>>
[2, 3, 3]
I wanna find fastest way to get expected result, not using for loop.
Would you suggest any good idea?
Edit:
Oops, Sorry for confusing you.
I saying 'not usting for loop' means, I wanna use list comprehension instead of for loop, because I heard list comprehension fater than for loop.
Use a nested list comprehension:
[j for i in ls for j in i if j != 0]
ls = [[i for i in np.random.randint(0, 5, 5)] for _ in range(7)]
[[1, 0, 3, 0, 0],
[1, 2, 2, 3, 0],
[1, 1, 1, 4, 3],
[1, 0, 3, 0, 4],
[2, 0, 3, 0, 2],
[1, 0, 4, 4, 0],
[2, 4, 1, 1, 2]]
[j for i in ls for j in i if j != 0]
# [1, 3, 1, 2, 2, 3, 1, 1, 1, 4, 3, 1, 3, 4, 2, 3, 2, 1, 4, 4, 2, 4, 1, 1, 2]
If you want to avoid any explicit looping here's an option using itertools.chain and filter:
from itertools import chain
list(filter(lambda x: x != 0, chain(*ls)))
# [1, 3, 1, 2, 2, 3, 1, 1, 1, 4, 3, 1, 3, 4, 2, 3, 2, 1, 4, 4, 2, 4, 1, 1, 2]
Looks like you're also using NumPy for creating the list. Note that this would be way simpler and more efficient using np.nonzero:
import numpy as np
a = np.random.randint(0, 5, (7,5))
a[np.nonzero(a)]
# [1, 3, 1, 2, 2, 3, 1, 1, 1, 4, 3, 1, 3, 4, 2, 3, 2, 1, 4, 4, 2, 4, 1, 1, 2]
Adding to #yatu's answer, since it's 0, you can just do if x:
print([x for i in ls for x in i if x])
Or without loop:
print(np.array(ls).flatten()[np.array(ls).flatten() != 0].tolist())
Both output:
[1, 4, 4, 2, 1, 3, 3, 4, 4, 1, 3, 3, 2, 2, 2, 1, 4, 2, 2, 1, 4, 1, 2, 3, 2, 4]
I'm trying to sort a list by moving through it and checking if a given element of the list is greater than the next element of the list, and if so, moving it accordingly so that the smaller number is to the left, larger to the right.
This is the code I have so far:
L = [3, 4, 1, 5, 2, 0]
for i in range(0, (len(L)-1)):
if L[i] > L[i+1]:
L[i], L[i+1] = L[i+1], L[i]
print(L)
The output of this for the three iterations are as follows:
[3, 1, 4, 5, 2, 0]
[3, 1, 4, 2, 5, 0]
[3, 1, 4, 2, 0, 5]
My goal is to get it to read [0, 1, 2, 3, 4, 5]. I realize I can just use the sorted() function but I'd prefer not to. I was thinking that I could try and iterate the for loop over the list multiple times, but I am unsure as to how to do so.
Currently you only compare each element and it's successor. However you really want to compare it to all following elements (starting with i+1 and ending with len(L)), so you need a double loop:
L = [3, 4, 1, 5, 2, 0]
for i in range(0, (len(L)-1)):
for j in range(i+1, len(L)):
if L[i] > L[j]:
L[i], L[j] = L[j], L[i]
print(L)
Which prints the following steps:
[1, 4, 3, 5, 2, 0]
[0, 4, 3, 5, 2, 1]
[0, 3, 4, 5, 2, 1]
[0, 2, 4, 5, 3, 1]
[0, 1, 4, 5, 3, 2]
[0, 1, 3, 5, 4, 2]
[0, 1, 2, 5, 4, 3]
[0, 1, 2, 4, 5, 3]
[0, 1, 2, 3, 5, 4]
[0, 1, 2, 3, 4, 5]
You could just put it inside a while loop:
L = [3, 4, 1, 5, 2, 0]
flag = True
while flag:
flag = False
for i in range(0, (len(L)-1)):
if L[i] > L[i+1]:
L[i], L[i+1] = L[i+1], L[i]
flag = True
print(L)
Output:
[3, 1, 4, 5, 2, 0]
[3, 1, 4, 2, 5, 0]
[3, 1, 4, 2, 0, 5]
[1, 3, 4, 2, 0, 5]
[1, 3, 2, 4, 0, 5]
[1, 3, 2, 0, 4, 5]
[1, 2, 3, 0, 4, 5]
[1, 2, 0, 3, 4, 5]
[1, 0, 2, 3, 4, 5]
[0, 1, 2, 3, 4, 5]