I have a list of tuples:
l=[(1,2,3),(4,5,6)]
The list can be of arbitrary length, as can the tuples. I'd like to convert this into a list or tuple of the elements, in the order they appear:
f=[1,2,3,4,5,6] # or (1,2,3,4,5,6)
If I know the at development time how many tuples I'll get back, I could just add them:
m = l[0] + l[1] # (1,2,3,4,5,6)
But since I don't know until runtime how many tuples I'll have, I can't do that. I feel like there's a way to use map to do this, but I can't figure it out. I can iterate over the tuples and add them to an accumulator, but that would create lots of intermediate tuples that would never be used. I could also iterate over the tuples, then the elements of the tuples, and append them to a list. This seems very inefficient. Maybe there's an even easier way that I'm totally glossing over. Any thoughts?
Chain them (only creates a generator instead of reserving extra memory):
>>> from itertools import chain
>>> l = [(1,2,3),(4,5,6)]
>>> list(chain.from_iterable(l))
[1, 2, 3, 4, 5, 6]
l = [(1, 2), (3, 4), (5, 6)]
print sum(l, ()) # (1, 2, 3, 4, 5, 6)
reduce(tuple.__add__, [(1,2,3),(4,5,6)])
tuple(i for x in l for i in x) # (1, 2, 3, 4, 5, 6)
Use the pythonic generator style for all of the following:
b=[(1,2,3),(4,5,6)]
list = [ x for x in i for i in b ] #produces a list
gen = ( x for x in i for i in b ) #produces a generator
tup = tuple( x for x in i for i in b ) #produces a tuple
print list
>> [1, 2, 3, 4, 5, 6]
>>> from itertools import chain
>>> l = [(1,2,3),(4,5,6)]
>>> list(chain(*l))
[1, 2, 3, 4, 5, 6]
You can combine the values in a list using the .extend() function like this:
l = [(1,2,3), (4,5,6)]
m = []
for t in l:
m.extend(t)
or a shorter version using reduce:
l = [(1,2,3), (4,5,6)]
m = reduce(lambda x,y: x+list(y), l, [])
Related
I have following objects
a = ("one", 1)
b = ("two", 2)
c = ("seven", 7)
my_dict = {10: a, 20:b, 70:c}
I want to get a list of [1, 2, 7] from my_dict. Whats the most pythonic way to do so?
Just use list comprehension. It's the most pythonic way of doing most things ;)
output = [x[1] for x in my_dict.values()]
If you like the functional way, you could use:
list(zip(*my_dict.values()))[1]
Output: (1, 2, 7)
Or, as list:
list(list(zip(*my_dict.values()))[1])
Output: [1, 2, 7]
You can use tuple unpacking inside the list comprehension to avoid having to index into the tuple:
result = [item for _, item in my_dict.values()]
print(result)
This prints:
[1, 2, 7]
I have a list:
list = [1, 2, 3]
I want to get this list and add a tuple so this list becomes like this
T_list = [(1,x), (2, x), (3,x)]
How do I do it?
Use a simple list comprehension to do this:
>>> your_list = [1, 2, 3]
>>> x = 100
>>> your_list_with_tuples = [(k, x) for k in your_list]
>>> your_list_with_tuples
Output
[(1, 100), (2, 100), (3, 100)]
Also, don't name your variables list, dict, etc, because they shadow the builtin types/functions.
A list comprehension would do the trick:
t_list = [(y,x) for y in original_list]
Also, the commenter is right. Don't name your lists list. It's a built in function. Using it for something else could cause problems down the line.
Here's my attempt, although I'm not an expert.
lis = [1,2,3]
tup = (4, 5, 6)
new_tup = tuple(lis)
new_lis = []
for i in range(0,3):
data = (new_tup[i],tup[i])
new_lis.append(data)
print(new_lis)
I have a question that I haven't quite found a good solution to. I'm looking for a better way to append function output to two or more lists, without using temp variables. Example below:
def f():
return 5,6
a,b = [], []
for i in range(10):
tmp_a, tmp_b = f()
a.append(tmp_a)
b.append(temp_b)
I've tried playing around with something like zip(*f()), but haven't quite found a solution that way.
Any way to remove those temp vars would be super helpful though, thanks!
Edit for additional info:
In this situation, the number of outputs from the function will always equal the number of lists that are being appended to. The main reason I'm looking to get rid of temps is for the case where there are maybe 8-10 function outputs, and having that many temp variables would get messy (though I don't really even like having two).
def f():
return 5,6
a,b = zip(*[f() for i in range(10)])
# this will create two tuples of elements 5 and 6 you can change
# them to list by type casting it like list(a), list(b)
First solution: we make a list of all results, then transpose it
def f(i):
return i, 2*i
# First make a list of all your results
l = [f(i) for i in range(5)]
# [(0, 0), (1, 2), (2, 4), (3, 6), (4, 8)]
# then transpose it using zip
a, b = zip(*l)
print(list(a))
print(list(b))
# [0, 1, 2, 3, 4]
# [0, 2, 4, 6, 8]
Or, all in one line:
a, b = zip(*[f(i) for i in range(5)])
A different solution, building the lists at each iteration, so that you can use them while they're being built:
def f(i):
return 2*i, i**2, i**3
doubles = []
squares = []
cubes = []
results = [doubles, squares, cubes]
for i in range(1, 4):
list(map(lambda res, val: res.append(val), results, f(i)))
print(results)
# [[2], [1], [1]]
# [[2, 4], [1, 4], [1, 8]]
# [[2, 4, 6], [1, 4, 9], [1, 8, 27]]
print(cubes)
# [1, 8, 27]
Note about list(map(...)): in Python3, map returns a generator, so we must use it if we want the lambda to be executed.list does it.
For your specific case, the zip answers are great.
Using itertools.cycle and itertools.chain is a different approach from the existing answers that might come in handy if you have a lot of pre-existing lists that you want to append to in a round-robin fashion. It also works when your function returns more values than you have lists.
>>> from itertools import cycle, chain
>>> a, b = [], [] # new, empty lists only for demo purposes
>>> for l, v in zip(cycle([a, b]), (chain(*(f() for i in range(10))))):
... l.append(v)
...
>>> a
[5, 5, 5, 5, 5, 5, 5, 5, 5, 5]
>>> b
[6, 6, 6, 6, 6, 6, 6, 6, 6, 6]
I'd do
tmp = f()
a.append(tmp[0])
b.append(tmp[1])
Not sure how pythonic it is for you though.
Good Day, I've googled this question and have found similar answers but not what I am looking for. I am not sure what the problem is called so I that doesn't help me and I am looking for an elegant solution.
How do I loop over a list, item at a time, and compare it to all other items in a list. For example, if I had a list
l = [1,2,3,4]
Each loop of the out would yield something like
1 vs [2,3,4]
2 vs [1,3,4]
3 vs [1,2,4]
4 vs [1,2,3]
One solution I've been playing with involves duplicating the list every iteration, finding the index of the item, deleting it from the duplicate list and compare the two. This route seems less ideal as you have to create a new list on every iteration.
You can use itertools.combiations to create all combinations of the length 3 from your list and then use set.defference method to get the difference element between the l and the combinations. but note that you need to convert your main list to a set object :
>>> from itertools import combinations
>>> l = {1,2,3,4}
>>> [(l.difference(i).pop(),i) for i in combinations(l,3)]
[(4, (1, 2, 3)), (3, (1, 2, 4)), (2, (1, 3, 4)), (1, (2, 3, 4))]
A simple approach would be to use two loops:
arr = [1,2,3,4]
for i in arr:
comp = []
for j in arr:
if i != j:
comp.append(j)
print(comp)
I guess you could use list comprehension. While still creating a new list every iteration, you don't need to delete an item each time:
l = [1,2,3,4]
for i in l:
temp = [item for item in l if item != i]
print temp
[2, 3, 4]
[1, 3, 4]
[1, 2, 4]
[1, 2, 3]
Let's see if I can explain what I am looking for. I want unwrap a list according to those elements which are a list in a list (tweaky, I know!).
I want this:
a = [[1],[2],[5],[4,6],[3]]
to be transform to this:
b = [[[1],[2],[5],[4],[3]],[[1],[2],[5],[6],[3]]]
And also applied to:
a = [[1],[2,3],[5],[4,6],[3]]
for achieving:
b = [[[1],[2],[5],[4],[3]],[[1],[2],[5],[6],[3]],[[1],[3],[5],[4],[3]],[[1],[3],[5],[6],[3]]]
I hope I have expressed propertly. I have been looking for some buil-in function that performs this operation but I have find nothing.
Thanks you in advance
You're looking for itertools.product:
>>> import itertools
>>> a = [[1],[2,3],[5],[4,6],[3]]
>>> for tup in itertools.product(*a):
... print tup
...
(1, 2, 5, 4, 3)
(1, 2, 5, 6, 3)
(1, 3, 5, 4, 3)
(1, 3, 5, 6, 3)
You can convert the tuples to lists and wrap the elements in 1-element lists if you really want to.