I have a nested list say:
lst = [[1,2,3,4], [2,3,4,5], [3,4,5,6]]
And I would like the output to be:
new_list = [[[1], [2], [3], [4]], [[2], [3], [4], [5]], [[3], [4], [5], [6]]]
this is what i am thinking, but I it is outputting a flat, list of list.
new_lst = []
for i in lst:
i = list(i)
for el in i:
new_list.append([el])
print(new_lst)
I would like to maintain the length of each predefined list in lst
Try List comprehension
[[ [e] for e in l] for l in lst]
You could use list comprehension and append every element to another list.
lst = [[1,2,3,4], [2,3,4,5], [3,4,5,6]]
new = [[[numbers] for numbers in elements] for elements in lst]
The above example adjusts for your desired output.
[[[1], [2], [3], [4]], [[2], [3], [4], [5]], [[3], [4], [5], [6]]]
You can use numpy's reshape np.reshape
>>> import numpy as np
>>> lst = np.array([[1,2,3,4], [2,3,4,5], [3,4,5,6]])
>>> lst
array([[1, 2, 3, 4],
[2, 3, 4, 5],
[3, 4, 5, 6]])
>>> lst.shape
(3, 4)
>>> lst.reshape(3,4,1)
array([[[1],
[2],
[3],
[4]],
[[2],
[3],
[4],
[5]],
[[3],
[4],
[5],
[6]]])
>>> lst.reshape(3,4,1).tolist()
[[[1], [2], [3], [4]], [[2], [3], [4], [5]], [[3], [4], [5], [6]]]
Another version, using recursion (you can have more level of depths in your input):
lst = [[1, 2, 3, 4], [2, 3, 4, 5], [3, 4, 5, 6]]
def split(i):
if isinstance(i, list):
return [split(v) for v in i]
else:
return [i]
print(split(lst))
Prints:
[[[1], [2], [3], [4]], [[2], [3], [4], [5]], [[3], [4], [5], [6]]]
You could use helper function to have more flexibility later of how you divide the smaller lists. based on an answer from here.
def split_list(alist, wanted_parts=1):
length = len(alist)
return [ alist[i*length // wanted_parts: (i+1)*length // wanted_parts]
for i in range(wanted_parts) ]
lst = [[1,2,3,4], [2,3,4,5], [3,4,5,6]]
ret =[]
for mini_list in lst:
ret.append(split_list(mini_list, len(mini_list)))
I think you had the right idea using multiple for loops. This should work:
list = [[1,2,3,4], [2,3,4,5], [3,4,5,6]]
for i in range(0, 3):
for j in range(0, 4):
(list[i])[j] = [(list[i])[j]]
Related
I have a list J with len(J)=2. I want to create a sublist of each element in J[i] where i=0,1. I present the current and expected output.
J = [[1, 2, 4, 6, 7],[1,4]]
arJ1=[]
for i in range(0,len(J)):
J1=[J[i]]
arJ1.append(J1)
J1=list(arJ1)
print("J1 =",J1)
The current output is
J1 = [[[1, 2, 4, 6, 7], [1, 4]]]
The expected output is
J1 = [[[1], [2], [4], [6], [7]], [[1], [4]]]
you can try this,
J = [[1, 2, 4, 6, 7],[1,4]]
new_l = []
for l in J:
tmp = []
for k in l:
tmp.append([k])
new_l.append(tmp)
print(new_l)
this will give you
[[[1], [2], [4], [6], [7]], [[1], [4]]]
With simple list comprehension:
res = [[[i] for i in sub_l] for sub_l in J]
print(res)
[[[1], [2], [4], [6], [7]], [[1], [4]]]
You can do in one line with list comprehension:
[[[k] for k in l] for l in J]
J = [[1, 2, 4, 6, 7],[1,4]]
arJ1 = []
for in_list in J:
new_list = []
for x in in_list:
new_list.append([x])
arJ1.append(new_list)
print(arJ1)
generate all combinations from list of numbers,the combinations can be pair of two numbers
example 1 : list of 2 numbers [1,2]
[
[[1],[2]],
[[1,2]]
]
example 2 : list of 3 numbers [1,2,3]
[
[[1], [2], [3]],
[[1], [2, 3]],
[[1, 3], [2]],
[[1, 2], [3]]
]
example 3 : list of 4 numbers [1,2,3,4]
[
[[1], [2], [3], [4]]
[[1], [2], [3, 4]],
[[1], [2, 4], [3]],
[[1], [2, 3], [4]],
[[1, 4], [2], [3]],
[[1, 3], [2], [4]],
[[1, 2], [3], [4]],
[[1, 2], [3, 4]],
[[1, 3], [2, 4]],
[[1, 4], [2, 3]]
]
The current implementation works but it is slow for list of 10 numbers
def get_all_order_combinations(nums, first=True):
if first and len(nums) == 1:
return [[[nums[0]]]]
if len(nums) == 2:
nums.sort()
return [
[[nums[0]], [nums[1]]],
[[nums[0], nums[1]]]
]
else:
all_results = []
for i in range(0, len(nums)):
temp_list = list(nums)
del temp_list[i]
current_num = nums[i]
results = get_all_order_combinations(temp_list, False)
results = [[[current_num]]+result for result in results]
for result in results:
result.sort()
if result not in all_results:
all_results.append(result)
if len(nums) >= 4:
for comb in combinations(nums, 2):
comb = list(comb)
results = get_all_order_combinations(
[n for n in nums if n not in comb]
,False
)
results = [[comb]+result for result in results]
for result in results:
result.sort()
if result not in all_results:
all_results.append(result)
return all_results
Try using the python in-built "itertools" package.
It is optimised to create combinations, which may solve your concerns with speed. Online searches will yield results on how to use it.
Here is one example:
geeksforgeeks Permutation and Combination in Python with itertools
My attempt at reproducing example 2 using itertools:
from itertools import combinations
num_list = [1, 2, 3]
output_list = []
for i in range(1, len(num_list)):
comb = combinations(num_list, i)
obtained_combinations = []
for combination in list(comb):
if i != 1:
temp_list = num_list.copy()
for selected_number in combination:
temp_list.remove(selected_number)
obtained_combinations.append([temp_list, list(combination)])
else:
obtained_combinations.append(list(combination))
output_list.append(obtained_combinations)
print(output_list)
Output:
[[[1], [2], [3]], [[[3], [1, 2]], [[2], [1, 3]], [[1], [2, 3]]]]
How can I take a nested List and group it into a tuple of three without importing any module. Please show the expanded for loops so I can understand it better.
For example, I want this nested List. Note this will always multiples of
3 sub lists so there is not a index error.
Thankyou.
[[1], [2], [3], [4], [5], [6], [7], [8], [9]] # Given
[(1, 2, 3), (4, 5, 6), (7, 8, 9)] # Result Wanted
No need for indexes. You can use next() on an iterator even inside a for loop:
xss = [[1], [2], [3], [4], [5], [6], [7], [8], [9]]
it = iter(xss)
answer = []
for x, in it:
answer.append((x, next(it)[0], next(it)[0]))
You can slice with a step size of 3 and zip to make the triples, do a nested unpacking in the for loop, and rebuild the triples without the wrapping lists.
xss = [[1], [2], [3], [4], [5], [6], [7], [8], [9]]
it = zip(xss[::3], xss[1::3], xss[2::3])
answer = []
for [x], [y], [z] in it:
answer.append((x, y, z))
given = [[1], [2], [3], [4], [5], [6], [7], [8], [9]] # because this is a list within a list
output = []
for i in range(0, len(given),3): # step size 3 indicated
temp = (given[i][0], given[i+1][0], given[i+2][0]) # therefore you need the "[0]" behind each given[i]
output.append(temp)
print (output)
I have a Python list and would like to build the ranged loop list. It looks like a ring buffer. So if I have a list:
[[0], [1], [2], [3]]
I would like to get:
[[0], [1], [2], [3]]
[[1], [2], [3], [4]]
[[2], [3], [4], [0]]
[[3], [4], [0], [1]]
[[4], [0], [1], [2]]
I could do it by myself. But is there any better or smarter ways in Python 3?
The code I have tried:
N = 5
d_list = [[_] for _ in range(N)]
for i in range(N):
b1 = i
e1 = i + N - 1
b2, e2 = 0, 0
if e1 >= N:
e2 = e1 - N
print(d_list[b1:e1] + d_list[b2:e2])
what about using a collections.deque and rotate ?
import collections
N = 5
d = collections.deque(range(N))
for _ in range(N):
print(d)
d.rotate(1)
result:
deque([0, 1, 2, 3, 4])
deque([4, 0, 1, 2, 3])
deque([3, 4, 0, 1, 2])
deque([2, 3, 4, 0, 1])
deque([1, 2, 3, 4, 0])
rotate just changes the start of the list, no data is copied/moved, so it's very fast.
note:
you can convert into list if needed
my example is using integers, not lists containing one sole integer. This can be easily adapted if necessary.
This Python function rotates anything slice-able as you desire:
def rotate(l, y=1):
if len(l) == 0:
return l
y = -y % len(l) # flip rotation direction
return l[y:] + l[:y]
>>> rotate([1,2,3,4,5],2)
[4, 5, 1, 2, 3]
>>> rotate([1,2,3,4,5],-22)
[3, 4, 5, 1, 2]
>>> rotate('abcdefg',3)
'efgabcd'
>>> for i in range(N):
... print(d_list)
... d_list=rotate(d_list)
...
[[0], [1], [2], [3], [4]]
[[4], [0], [1], [2], [3]]
[[3], [4], [0], [1], [2]]
[[2], [3], [4], [0], [1]]
[[1], [2], [3], [4], [0]]
Note that in Python 3, range is not sliceable. You first would need to create a list with list(range(...))
The sign convention is the same as deque.rotate
This follows on from this question:
Algorithm to generate spanning set
Given this input: [1,2,3,4]
I'd like to generate this set of sets in python:
[1] [2] [3] [4]
[1] [2] [3,4]
[1] [2, 3, 4]
[1] [2,3] [4]
[1,2] [3] [4]
[1,2] [3,4]
[1,2,3] [4]
[1,2,3,4]
So unlike the previous question, the order of the list is retained.
Ideally the code would work for n items in the list
Thanks very much
EDIT 2: Could anyone advise me on how to do this if the original input is a string rather than a list (where each word in the string becomes an item in a list). Thanks!
EDIT: added [1] [2, 3, 4] Sorry for the mistake
You might also enjoy a recursive solution:
def span(lst):
yield [lst]
for i in range(1, len(lst)):
for x in span(lst[i:]):
yield [lst[:i]] + x
Explanation
We exploit recursion here to break the problem down. The approach is the following:
For every list, the whole list is a valid spanning: [1,2,3,4] => [[1,2,3,4]].
For every list that is longer than size 1, we can use the first item as a group and then apply the same algorithm on the remaining list to get all the combined results:
[1,2,3] =>
[[1]] + [[2], [3]] # => [[1], [2], [3]]
[[1]] + [[2,3]] # => [[1], [2,3]]
For every list that is longer than size 2, we can just as well use the first two items as a group and then apply the same algorithm on the remaining list and combine the results:
[1,2,3,4,5] =>
[[1,2]] + [[3], [4], [5]] # => [[1,2], [3], [4], [5]]
[[1,2]] + [[3,4], [5]] # => [[1,2], [3,4], [5]]
[[1,2]] + [[3], [4,5]] # => [[1,2], [3], [4,5]]
[[1,2]] + [[3,4,5]] # => [[1,2], [3,4,5]]
We can see that the possible combinations on the right side are indeed all possible groupings of the remainder of the list, [3,4,5].
For every list that is longer than ... etc. Thus, the final algorithm is the following:
yield the whole list (it is always a valid spanning, see above)
For every possible splitting of the list, yield the left-hand part of the list combined with all possible spannings of the right-hand part of the list.
yield is a special keyword in Python that make the function a generator, which means that it returns a iterable object that can be used to enumerate all results found. You can transform the result into a list using the list constructor function: list(span([1,2,3,4])).
Adjusting one of the solution from Python: show all possible groupings of a list:
from itertools import combinations
def cut(lst, indexes):
last = 0
for i in indexes:
yield lst[last:i]
last = i
yield lst[last:]
def generate(lst, n):
for indexes in combinations(list(range(1,len(lst))), n - 1):
yield list(cut(lst, indexes))
data = [1,2,3,4]
for i in range(1, len(data)+1): # the only difference is here
for g in generate(data, i):
print(g)
"""
[[1, 2, 3, 4]]
[[1], [2, 3, 4]]
[[1, 2], [3, 4]]
[[1, 2, 3], [4]]
[[1], [2], [3, 4]]
[[1], [2, 3], [4]]
[[1, 2], [3], [4]]
[[1], [2], [3], [4]]
"""
import itertools
a = [1, 2, 3, 4]
n = len(a)
for num_splits in range(n):
for splits in itertools.combinations(range(1, n), num_splits):
splices = zip([0] + list(splits), list(splits) + [n])
print([a[i:j] for i, j in splices])
prints
[[1, 2, 3, 4]]
[[1], [2, 3, 4]]
[[1, 2], [3, 4]]
[[1, 2, 3], [4]]
[[1], [2], [3, 4]]
[[1], [2, 3], [4]]
[[1, 2], [3], [4]]
[[1], [2], [3], [4]]