I am trying to group list of items relevant to a query item. Below is an example of the problem and my attempt at it:
>>> _list=[[1,2,3],[2,3,4]]
>>> querylist=[1,2,4]
>>> relvant=[]
>>> for x in querylist:
for y in _list:
if x in y:
relvant.append(y)
My output:
>>> relvant
[[1, 2, 3], [1, 2, 3], [2, 3, 4], [2, 3, 4]]
Desired output:
[[[1, 2, 3]], [[1, 2, 3], [2, 3, 4]],[[2, 3, 4]]]
The issue is after each loop of a query item, I expected the relevant lists to be grouped but that isn't the case with my attempt.
Thanks for your suggestions.
I think it's clearer to use a list comprehension:
>>> _list = [[1,2,3],[2,3,4]]
>>> querylist = [1,2,4]
>>> [[l for l in _list if x in l] for x in querylist]
[[[1, 2, 3]], [[1, 2, 3], [2, 3, 4]], [[2, 3, 4]]]
The inner expression [l for l in _list if x in l] describes the list of all sublists that contain x. The outer expression's job is to get that list for all values of x in the query list.
By making minimal changes in the code provided you can create new dummy list to store values and at end of each inner loop iteration you just append it to the main list.
_list=[[1,2,3],[2,3,4]]
querylist=[1,2,4]
relvant=[]
for x in querylist:
dummy = []
for y in _list:
if x in y:
dummy.append(y)
relvant.append(dummy)
print relvant
>>> [[[1, 2, 3]], [[1, 2, 3], [2, 3, 4]],[[2, 3, 4]]]
Related
I have a dictionary, each key of dictionary has a list of list (nested list) as its value. What I want is imagine we have:
x = {1: [[1, 2], [3, 5]], 2: [[2, 1], [2, 6]], 3: [[1, 5], [5, 4]]}
My question is how can I access each element of the dictionary and concatenate those with same index: for example first list from all keys:
[1,2] from first keye +
[2,1] from second and
[1,5] from third one
How can I do this?
You can access your nested list easily when you're iterating through your dictionary and append it to a new list and the you apply the sum function.
Code:
x={1: [[1,2],[3,5]] , 2:[[2,1],[2,6]], 3:[[1,5],[5,4]]}
ans=[]
for key in x:
ans += x[key][0]
print(sum(ans))
Output:
12
Assuming you want a list of the first elements, you can do:
>>> x={1: [[1,2],[3,5]] , 2:[[2,1],[2,6]], 3:[[1,5],[5,4]]}
>>> y = [a[0] for a in x.values()]
>>> y
[[1, 2], [2, 1], [1, 5]]
If you want the second element, you can use a[1], etc.
The output you expect is not entirely clear (do you want to sum? concatenate?), but what seems clear is that you want to handle the values as matrices.
You can use numpy for that:
summing the values
import numpy as np
sum(map(np.array, x.values())).tolist()
output:
[[4, 8], [10, 15]] # [[1+2+1, 2+1+5], [3+2+5, 5+6+4]]
concatenating the matrices (horizontally)
import numpy as np
np.hstack(list(map(np.array, x.values()))).tolist()
output:
[[1, 2, 2, 1, 1, 5], [3, 5, 2, 6, 5, 4]]
As explained in How to iterate through two lists in parallel?, zip does exactly that: iterates over a few iterables at the same time and generates tuples of matching-index items from all iterables.
In your case, the iterables are the values of the dict. So just unpack the values to zip:
x = {1: [[1, 2], [3, 5]], 2: [[2, 1], [2, 6]], 3: [[1, 5], [5, 4]]}
for y in zip(*x.values()):
print(y)
Gives:
([1, 2], [2, 1], [1, 5])
([3, 5], [2, 6], [5, 4])
How can I "pack" consecutive duplicated elements in a list into sublists of the repeated element?
What I mean is:
l = [1, 1, 1, 2, 2, 3, 4, 4, 1]
pack(l) -> [[1,1,1], [2,2], [3], [4, 4], [1]]
I want to do this problem in a very basic way as I have just started i.e using loops and list methods. I have looked for other methods but they were difficult for me to understand
For removing the duplicates instead of packing them, see Removing elements that have consecutive duplicates
You can use groupby:
from itertools import groupby
def pack(List):
result = []
for key, group in groupby(List):
result.append(list(group))
return result
l = [1, 1, 1, 2, 2, 3, 4, 4, 1]
print(pack(l))
Or one-line:
l = [1, 1, 1, 2, 2, 3, 4, 4, 1]
result = [list(group) for key,group in groupby(l)]
# [[1, 1, 1], [2, 2], [3], [4, 4], [1]]
You can use:
lst = [1, 1, 1, 2, 2, 3, 4, 4, 1]
# bootstrap: initialize a sublist with the first element of lst
out = [[lst[0]]]
for it1, it2 in zip(lst, lst[1:]):
# if previous item and current one are equal, append result to the last sublist
if it1 == it2:
out[-1].append(it2)
# else append a new empty sublist
else:
out.append([it2])
Output:
>>> out
[[1, 1, 1], [2, 2], [3], [4, 4], [1]]
This code will do:
data = [0,0,1,2,3,4,4,5,6,6,6,7,8,9,4,4,9,9,9,9,9,3,3,2,45,2,11,11,11]
newdata=[]
for i,l in enumerate(data):
if i==0 or l!=data[i-1]:
newdata.append([l])
else:
newdata[-1].append(l)
#Output
[[0,0],[1],[2],[3],[4,4],[5],[6,6,6],[7],[8],[9],[4,4],[9,9,9,9,9],[3,3],[2],[45],[2],[11,11,11]]
How can i create a new list combing the first values of my old lists and then the second ones etc..
list_1 = [1,2,3,4]
list_2 = [1,2,3,4]
list_3 = [1,2,3,4]
new_list = [[1,1,1],[2,2,2],[3,3,3],[4,4,4]]
Pure python:
You can use zip:
new_list = list(map(list, zip(list_1,list_2,list_3)))
>>> new_list
[[1, 1, 1], [2, 2, 2], [3, 3, 3], [4, 4, 4]]
Alternative:
numpy:
import numpy as np
new_list = np.array([list_1,list_2,list_3]).T.tolist()
>>> new_list
[[1, 1, 1], [2, 2, 2], [3, 3, 3], [4, 4, 4]]
Here's another way to do it using list comprehensions.
new_list = [list(args) for args in zip(list_1, list_2, list_3)]
If we enumerate one list the index from that list to all 3
new_list = [[list_1[i], list_2[i], list_3[i]] for i, _ in enumerate(list_1)]
# [[1, 1, 1], [2, 2, 2], [3, 3, 3], [4, 4, 4]]
I got a list
a=[1,2,3]
and a list of list
b=[[1,2],[3,4,5]]
and I want to insert a into b at index 1 so b becomes
b=[[1,2],[1,2,3],[3,4,5]]
How do I do that?If I use insert it won't work because I can only insert an item not a list?
EDIT:I realised insert can be used for lists as well.Thanks.
You can use list.insert which takes the index as the first argument
>>> a=[1,2,3]
>>> b=[[1,2],[3,4,5]]
>>> b.insert(1, a)
>>> b
[[1, 2], [1, 2, 3], [3, 4, 5]]
You can use list slicing:
b=[[1,2],[3,4,5]]
a = [1, 2, 3]
final_list = b[:1]+[a]+b[1:]
Output:
[[1, 2], [1, 2, 3], [3, 4, 5]]
I have a list of lists representing a connectivity graph in Python. This list look like a n*2 matrix
example = [[1, 2], [1, 5], [1, 8], [2, 1], [2, 9], [2,5] ]
what I want to do is to find the value of the first elements of the lists where the second element is equal to a user defined value. For instance :
input 1 returns [2] (because [2,1])
input 5 returns [1,2] (because [1,5] and [2,5])
input 7 returns []
in Matlab, I could use
output = example(example(:,1)==input, 2);
but I would like to do this in Python (in the most pythonic and efficient way)
You can use list comprehension as a filter, like this
>>> example = [[1, 2], [1, 5], [1, 8], [2, 1], [2, 9], [2,5]]
>>> n = 5
>>> [first for first, second in example if second == n]
[1, 2]
You can work with the Python functions map and filter very comfortable:
>>> example = [[1, 2], [1, 5], [1, 8], [2, 1], [2, 9], [2,5] ]
>>> n = 5
>>> map(lambda x: x[0], filter(lambda x: n in x, example))
[1,2]
With lambda you can define anonyme functions...
Syntax:
lambda arg0,arg1...: e
arg0,arg1... are your parameters of the fucntion, and e is the expression.
They use lambda functions mostly in functions like map, reduce, filter etc.
exemple = [[1, 2], [1, 5], [1, 8], [2, 1], [2, 9], [2,5] ]
foundElements = []
** input = [...] *** List of Inputs
for item in exemple:
if item[1] in input :
foundElements.append(item[0])