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I want to replace variables in a list with combination of elements.
To be more specific:
I have these two lists
liste1 = [1,2,3,'X','X',4]
liste2 = [5,6,7]
and I want to get a list containing the elements below :
[1,2,3,5,6,4]
[1,2,3,5,7,4]
[1,2,3,6,7,4]
[1,2,3,6,5,4]
[1,2,3,7,5,4]
[1,2,3,7,6,4]
Does anyone have an idea how to make it ?
You can do it this way:
from itertools import permutations
liste1 = [1, 2, 3, 'X', 'X', 4]
liste2 = [5, 6, 7]
def replacements(liste1, liste2):
x_indices = [i for i, val in enumerate(liste1) if val == 'X']
nb = len(x_indices)
for perm in permutations(liste2, nb):
l1 = liste1[:] # if we want to preserve the original and yield different lists
for i, new_val in zip(x_indices, perm):
l1[i] = new_val
yield l1
for r in replacements(liste1, liste2):
print(r)
Output:
[1, 2, 3, 5, 6, 4]
[1, 2, 3, 5, 7, 4]
[1, 2, 3, 6, 5, 4]
[1, 2, 3, 6, 7, 4]
[1, 2, 3, 7, 5, 4]
[1, 2, 3, 7, 6, 4]
We first list the indices where 'X' appears, then generate the permutations of as many elements of liste2. For each permutation, we replace the 'X's.
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Suppose I have a python list as: points_list = [1, 2, 3, 4, 5, 6, 7, 8, 9,1]
And I need to split this list containing the number of elements in the index_list=[2, 2, 6, 3] But having common endpoints
That is:
First 2 elements from points_list : [1,2]
Next 2 elements from the points_list but it should start from the place it stopped before : [2,3]
Then the next 6 elements : [3,4,5,6,7,8]
Finally, 3 elements in the same way : [8,9,1]
Ultimately, what I expect is to have something like: [[1,2],[2,3],[3,4,5,6,7,8],[8,9,1]] which corresponds to the number of elements mentioned in the index_list=[2, 2, 6, 3]
Can you please help me to perform this task
Here's my code:
points_list = [1, 2, 3, 4, 5, 6, 7, 8, 9,1]
index_list=[2, 2, 6, 3]
res = []
end_point = 0
for i in index_list:
temp_list = points_list[end_point:end_point+i] # Get the list
res.append(temp_list) # Append it
end_point = points_list.index(temp_list[-1]) # Get the index of the last value
print(res)
Support for repeating numbers (thanks to azro and S.B):
points_list = [3, 2, 3, 4, 5, 6, 2, 8, 9, 1]
index_list=[2, 2, 6, 3]
res = []
for i in index_list:
res.append(points_list[:i]) # Append it
points_list = points_list[i-1:] # Reconfigure list
print(res)
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So I'm trying to find the items that went in and went out of a list before and after it changed
For example, at first my list is:
[1, 1, 2, 5, 7, 7]
And then a minute later it's:
[1, 2, 2, 5, 6, 7, 4]
How would I end up with 2 lists that show what items went out and which went in like so,
itemsOut = [1,7]
itemsIn = [2, 6, 4]
Position and length can change
You can use Counter from the built-in collections module:
>>> list1 = [1, 1, 2, 5, 7, 7]
>>> list2 = [1, 2, 2, 5, 6, 7, 4]
>>> from collections import Counter
>>> counter1 = Counter(list1)
>>> counter2 = Counter(list2)
>>> l_diff = counter1-counter2
>>> r_diff = counter2-counter1
>>> print(list(l_diff))
[1, 7]
>>> print(list(r_diff))
[2, 6, 4]
You could use something like this, to detect wich items changed.
old_arr = [1, 2, 2, 5, 7, 7]
new_arr = [1, 1, 2, 5, 7, 7]
items_out = []
for element in old_arr:
if new_arr.count(element) > 0:
new_arr.pop(new_arr.index(element))
else:
items_out.append(element)
print("Items out:", items_out)
print("Items in:", new_arr)
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I am trying to count how many numbers appear in each row in a 2d list and then return the values with a new variable . For example, I would to like to return the items that are greater than or equal to that length.
Here my list below
df = [[2,4,6,7],[3,4,],[2,4,6,8,12,24],[3,5,7,333,450],[4,20]]
I would like to create a new variable and return each row that is longer than 3.
df2 =[[2,4,6,7],[2,4,6,8,12,24],[3,5,7,333,450]]
I would like to create a new variable and return each row with exactly 2 elements.
df3 = [[3,4,],[4,20]]
Probably not the most efficient solution but a succinct way to do this:
df = [[2,4,6,7],[3,4,],[2,4,6,8,12,24],[3,5,7,333,450],[4,20]]
df3 = filter(lambda l: len(l) > 3, df)
df2 = filter(lambda l: len(l) == 2, df)
print(list(df2))
print(list(df3))
output:
[[3, 4], [4, 20]]
[[2, 4, 6, 7], [2, 4, 6, 8, 12, 24], [3, 5, 7, 333, 450]]
You can use list comprehensions to do this. In this case, you can check the len of each sublist, and use that to filter out which sublists you want to keep/discard.
>>> df = [[2,4,6,7],[3,4,],[2,4,6,8,12,24],[3,5,7,333,450],[4,20]]
>>> [i for i in df if len(i) > 3]
[[2, 4, 6, 7], [2, 4, 6, 8, 12, 24], [3, 5, 7, 333, 450]]
>>> [i for i in df if len(i) == 2]
[[3, 4], [4, 20]]
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I have two lists say:
a = [1, 2, 2, 2, 3]
b = [2, 5, 6]
After doing a union, I should get something like this (don't mind the order):
c = [1, 2, 2, 2, 3, 5, 6]
The final list should contain common elements only once and rest of elements (from both lists) should be copied as they are. Sets can't be used for this as they remove multiple occurrence of an element from the list, which isn't a union. What is a Pythonic way to do so?
Perform a union, keeping repetition:
>>> c = a + b
[1, 2, 2, 3, 2, 5, 6]
Perform a union, keeping repetition & order:
>>> c = sorted(a + b)
[1, 2, 2, 2, 3, 5, 6]
Perform an union, no repetition in each list, but repetition allowed in final union, and keeped order:
>>> c = sorted(list(set(a)) + list(set(b)))
[1, 2, 2, 3, 5, 6]
After clarification of the question, the goal is to build a list that take elements (including repetition) of, and then add elements of b if they are not in the new list.
>>> c = a + [e for e in b if e not in a]
[1, 2, 2, 2, 3, 5, 6]
After another clarification of the question, the goal is to build a list containing all elements of input list. But, if elements are in common, they are pushed the same number there
>>> from collections import Counter
>>> def merge(a,b):
... na, nb = Counter(a), Counter(b)
... return list(Counter({k: max((na[k], nb[k])) for k in set(a + b)}).elements())
>>> merge([1, 2, 2, 2, 3], [2, 5, 6])
[1, 2, 2, 2, 3, 5, 6]
>>> merge([1, 2, 3], [2, 2, 4])
[1, 2, 2, 4]
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I have this function that returns the list of values and counts for integer values. I would like to alter this to return the counts and values of strings, as I am having trouble implementing. It works for integer values but not strings.
def frequencies(xs):
ys = sorted(xs)
values, count = [], []
for y in ys:
if y not in values:
values.append(y)
count.append(1)
else:
count[-1] += 1
return values, count
using itertools.groupby:
>>> import itertools
>>> a = [1,2,3,4,5,6,2,3,4,5,3,7,8,4,5,6,3]
>>> [ [x,len(list(y))] for x,y in itertools.groupby(sorted(a))] # if you want list
[[1, 1], [2, 2], [3, 4], [4, 3], [5, 3], [6, 2], [7, 1], [8, 1]]
>>> {x:len(list(y)) for x,y in itertools.groupby(sorted(a))} # if you want dictionary
{1: 1, 2: 2, 3: 4, 4: 3, 5: 3, 6: 2, 7: 1, 8: 1}
Why reinvent the wheel, and not use Counter? From what I understand, you try to achieve exactly what Counter does.
from collections import Counter
a = Counter([1,2,3,2,3,4,5,3,4,3,4,2,1,2,5])
print(a.keys(), a.values())
b = Counter(["1","2","3","2","3","4","5","3","4",'3','4','2','1','2','5'])
print(b.keys(), b.values())
Gives:
dict_keys([1, 2, 3, 4, 5]) dict_values([2, 4, 4, 3, 2])
dict_keys(['4', '2', '3', '1', '5']) dict_values([3, 4, 4, 2, 2])