Split list at a specific value - python

I am trying to write a code that splits lists in a class of lists in two when a certain value is a middle element of the list and then produce two lists where the middle element becomes the end element in the first list and the first element in the second one.
There can be more than n middle elements in the list so the result must be n+1 lists.
Example:
A = [[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15],[16,17,18,19,20,21,22,23,24,25],[26,27,28,29]]
P = [4,7,13,20]
n = len(Points) # in this case n = 4
I am looking for a result that looks like this:
A = [[0,1,2,3,4],[4,5,6,7],[7,8,9,10,11,12,13],[13,14,15],[16,17,18,19,20],[20,21,22,23,24,25],[26,27,28,29]]
Since n = 4 and it will produce 5 lists, note that the answer has 6 lists because the last list doesn't have any value of P in and therefore stays intact.
I haven't been able to produce anything as I am new to python and it is hard to formulate this problem.
Any help is appreciated!

You can first recover all indices of the provided values and then slice accordingly.
Code
def split_at_values(lst, values):
indices = [i for i, x in enumerate(lst) if x in values]
for start, end in zip([0, *indices], [*indices, len(lst)]):
yield lst[start:end+1]
# Note: remove +1 for separator to only appear in right side slice
Example
values = {4, 7, 13, 20}
lst = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
print(*split_at_values(lst, values))
Output
[0, 1, 2, 3, 4] [4, 5, 6, 7] [7, 8, 9, 10, 11, 12, 13] [13, 14, 15]
You can then apply this iteratively to you input list A to get the desired result. Alternatively you can use itertools.chain.from_iterable.
from itertools import chain
values = {4, 7, 13, 20}
lst_A = [[0, 1, 2, 3, 4, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[16, 17, 18, 19, 20, 21, 22, 23, 24, 25],
[26, 27, 28, 29]]
output = list(chain.from_iterable(split_at_values(sublst, values) for sublst in lst_A))
print(output)
Output
[[0, 1, 2, 3, 4],
[4, 5, 6, 7],
[7, 8, 9, 10, 11, 12, 13],
[13, 14, 15],
[16, 17, 18, 19, 20],
[20, 21, 22, 23, 24, 25],
[26, 27, 28, 29]]

You can keep appending the sub-list items to the last sub-list of the output list, and if the current item is equal to the next item in Points, append a new sub-list to the output with the same item and pop the item from Points:
output = []
for l in List:
output.append([])
for i in l:
output[-1].append(i)
if Points and i == Points[0]:
output.append([i])
Points.pop(0)
With your sample input, output would become:
[[0, 1, 2, 3, 4], [4, 5, 6, 7], [7, 8, 9, 10, 11, 12, 13], [13, 14, 15], [16, 17, 18, 19, 20], [20, 21, 22, 23, 24, 25], [26, 27, 28, 29]]

Related

How do I put data in a two-dimensional array sequentially without duplication?

I want to put data consisting of a one-dimensional array into a two-dimensional array. I will assume that the number of rows and columns is 5.
The code I tried is as follows.
data = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
a = []
for i in range(5):
a.append([])
for j in range(5):
a[i].append(j)
print(a)
# result : [[0, 1, 2, 3, 4], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4]]
# I want this : [[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20]]
You don't have to worry about the last [20].
The important thing is that the row must change without duplicating the data.
I want to solve it, but I can't think of any way. I ask for your help.
There are two issues with the current code.
It doesn't actually use any of the values from the variable data.
The data does not contain enough items to populate a 5x5 array.
After adding 0 to the beginning of the variable data and using the values from the variable, the code becomes
data = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
a = []
for i in range(5):
a.append([])
for j in range(5):
if i*5+j >= len(data):
break
a[i].append(data[i*5+j])
print(a)
The output of the new code will be
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20]]
This should deliever the desired output
data = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
a = []
x_i = 5
x_j = 5
for i in range(x_i):
a.append([])
for j in range(x_j):
a[i].append(i*x_j+j)
print(a)
Output:
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20, 21, 22, 23, 24]]
By using list comprehension...
data = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
columns = 5
rows = 5
result = [data[i * columns: (i + 1) * columns] for i in range(rows)]
print(result)
# [[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20]]
You could use itertools.groupby with an integer division to create the groups
from itertools import groupby
data = [0, 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
grouped_data = [list(v) for k, v in groupby(data, key=lambda x: x//5)]
print(grouped_data)
Output
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20]]

Turn 1D list to a triangle 2D list

I have a list made of range from 1 to 55 with step 2: [1, 3, 5, 7, 9, 11, 13, 15, ..., 53].
What I'm trying to do is to fill another list, which is 2D and triangular, with numbers from the first list.
def odd_triangle(n):
a = []
b = []
for number in range(1, 55, 2):
a.append(number)
for i in range(n):
b.append([])
for j in range(i+1):
b[i].append(a[i])
print(b)
After I call that function, for example odd_triangle(5) (5 rows will be created), it gives me not exactly what I want to:
[[1], [3, 3], [5, 5, 5], [7, 7, 7, 7], [9, 9, 9, 9, 9]]
What I want it to be is: [[1], [3, 5], [7, 9, 11], [13, 15, 17, 19], [21, 23, 25, 27, 29]]
This will work:
a = list(range(1,55,2))
n = 5
it = iter(a)
b = list([next(it) for _ in range(i)] for i in range(1, n+1))
Gives:
[[1], [3, 5], [7, 9, 11], [13, 15, 17, 19], [21, 23, 25, 27, 29]]
Here, next(it) simply gets the next value from the iterator over a every time it is called.

List comprehension function for odd-even numbers

I am trying to print a list containing 2 lists at index 0 and 1. One list contains even numbers and the other one odd numbers.
Also, I want to do it with list comprehension and use only one list variable.
even_odd = [[],[]]
even_odd = [even_odd[0].append(a) if a%2 == 0 else even_odd[1].append(a) for a in range(20)]
Expected Output:
[[0, 2, 4, 6, 8, 10, 12, 14, 16, 18], [1, 3, 5, 7, 9, 11, 13, 15, 17, 19]]
Using list comprehension
You can do it with two range by iterating in 2 interval
even_odd = [list(range(0, 19, 2)), list(range(1, 20, 2))]
# [[0, 2, 4, 6, 8, 10, 12, 14, 16, 18], [1, 3, 5, 7, 9, 11, 13, 15, 17, 19]]

Generating list of lists with custom value limitations with Hypothesis

The Story:
Currently, I have a function-under-test that expects a list of lists of integers with the following rules:
number of sublists (let's call it N) can be from 1 to 50
number of values inside sublists is the same for all sublists (rectangular form) and should be >= 0 and <= 5
values inside sublists cannot be more than or equal to the total number of sublists. In other words, each value inside a sublist is an integer >= 0 and < N
Sample valid inputs:
[[0]]
[[2, 1], [2, 0], [3, 1], [1, 0]]
[[1], [0]]
Sample invalid inputs:
[[2]] # 2 is more than N=1 (total number of sublists)
[[0, 1], [2, 0]] # 2 is equal to N=2 (total number of sublists)
I'm trying to approach it with property-based-testing and generate different valid inputs with hypothesis library and trying to wrap my head around lists() and integers(), but cannot make it work:
the condition #1 is easy to approach with lists() and min_size and max_size arguments
the condition #2 is covered under Chaining strategies together
the condition #3 is what I'm struggling with - cause, if we use the rectangle_lists from the above example, we don't have a reference to the length of the "parent" list inside integers()
The Question:
How can I limit the integer values inside sublists to be less than the total number of sublists?
Some of my attempts:
from hypothesis import given
from hypothesis.strategies import lists, integers
#given(lists(lists(integers(min_value=0, max_value=5), min_size=1, max_size=5), min_size=1, max_size=50))
def test(l):
# ...
This one was very far from meeting the requirements - list is not strictly of a rectangular form and generated integer values can go over the generated size of the list.
from hypothesis import given
from hypothesis.strategies import lists, integers
#given(integers(min_value=0, max_value=5).flatmap(lambda n: lists(lists(integers(min_value=1, max_value=5), min_size=n, max_size=n), min_size=1, max_size=50)))
def test(l):
# ...
Here, the #1 and #2 are requirements were being met, but the integer values can go larger than the size of the list - requirement #3 is not met.
There's a good general technique that is often useful when trying to solve tricky constraints like this: try to build something that looks a bit like what you want but doesn't satisfy all the constraints and then compose it with a function that modifies it (e.g. by throwing away the bad bits or patching up bits that don't quite work) to make it satisfy the constraints.
For your case, you could do something like the following:
from hypothesis.strategies import builds, lists, integers
def prune_list(ls):
n = len(ls)
return [
[i for i in sublist if i < n][:5]
for sublist in ls
]
limited_list_strategy = builds(
prune_list,
lists(lists(integers(0, 49), average_size=5), max_size=50, min_size=1)
)
In this we:
Generate a list that looks roughly right (it's a list of list of integers and the integers are in the same range as all possible indices that could be valid).
Prune out any invalid indices from the sublists
Truncate any sublists that still have more than 5 elements in them
The result should satisfy all three conditions you needed.
The average_size parameter isn't strictly necessary but in experimenting with this I found it was a bit too prone to producing empty sublists otherwise.
ETA: Apologies. I've just realised that I misread one of your conditions - this doesn't actually do quite what you want because it doesn't ensure each list is the same length. Here's a way to modify this to fix that (it gets a bit more complicated, so I've switched to using composite instead of builds):
from hypothesis.strategies import composite, lists, integers, permutations
#composite
def limisted_lists(draw):
ls = draw(
lists(lists(integers(0, 49), average_size=5), max_size=50, min_size=1)
)
filler = draw(permutations(range(50)))
sublist_length = draw(integers(0, 5))
n = len(ls)
pruned = [
[i for i in sublist if i < n][:sublist_length]
for sublist in ls
]
for sublist in pruned:
for i in filler:
if len(sublist) == sublist_length:
break
elif i < n:
sublist.append(i)
return pruned
The idea is that we generate a "filler" list that provides the defaults for what a sublist looks like (so they will tend to shrink in the direction of being more similar to eachother) and then draw the length of the sublists to prune to to get that consistency.
This has got pretty complicated I admit. You might want to use RecursivelyIronic's flatmap based version. The main reason I prefer this over that is that it will tend to shrink better, so you'll get nicer examples out of it.
You can also do this with flatmap, though it's a bit of a contortion.
from hypothesis import strategies as st
from hypothesis import given, settings
number_of_lists = st.integers(min_value=1, max_value=50)
list_lengths = st.integers(min_value=0, max_value=5)
def build_strategy(number_and_length):
number, length = number_and_length
list_elements = st.integers(min_value=0, max_value=number - 1)
return st.lists(
st.lists(list_elements, min_size=length, max_size=length),
min_size=number, max_size=number)
mystrategy = st.tuples(number_of_lists, list_lengths).flatmap(build_strategy)
#settings(max_examples=5000)
#given(mystrategy)
def test_constraints(list_of_lists):
N = len(list_of_lists)
# condition 1
assert 1 <= N <= 50
# Condition 2
[length] = set(map(len, list_of_lists))
assert 0 <= length <= 5
# Condition 3
assert all((0 <= element < N) for lst in list_of_lists for element in lst)
As David mentioned, this does tend to produce a lot of empty lists, so some average size tuning would be required.
>>> mystrategy.example()
[[24, 6, 4, 19], [26, 9, 15, 15], [1, 2, 25, 4], [12, 8, 18, 19], [12, 15, 2, 31], [3, 8, 17, 2], [5, 1, 1, 5], [7, 1, 16, 8], [9, 9, 6, 4], [22, 24, 28, 16], [18, 11, 20, 21], [16, 23, 30, 5], [13, 1, 16, 16], [24, 23, 16, 32], [13, 30, 10, 1], [7, 5, 14, 31], [31, 15, 23, 18], [3, 0, 13, 9], [32, 26, 22, 23], [4, 11, 20, 10], [6, 15, 32, 22], [32, 19, 1, 31], [20, 28, 4, 21], [18, 29, 0, 8], [6, 9, 24, 3], [20, 17, 31, 8], [6, 12, 8, 22], [32, 22, 9, 4], [16, 27, 29, 9], [21, 15, 30, 5], [19, 10, 20, 21], [31, 13, 0, 21], [16, 9, 8, 29]]
>>> mystrategy.example()
[[28, 18], [17, 25], [26, 27], [20, 6], [15, 10], [1, 21], [23, 15], [7, 5], [9, 3], [8, 3], [3, 4], [19, 29], [18, 11], [6, 6], [8, 19], [14, 7], [25, 3], [26, 11], [24, 20], [22, 2], [19, 12], [19, 27], [13, 20], [16, 5], [6, 2], [4, 18], [10, 2], [26, 16], [24, 24], [11, 26]]
>>> mystrategy.example()
[[], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], []]
>>> mystrategy.example()
[[], [], [], [], [], [], [], [], [], [], [], [], [], [], []]
>>> mystrategy.example()
[[6, 8, 22, 21, 22], [3, 0, 24, 5, 18], [16, 17, 25, 16, 11], [2, 12, 0, 3, 15], [0, 12, 12, 12, 14], [11, 20, 6, 6, 23], [5, 19, 2, 0, 12], [16, 0, 1, 24, 10], [2, 13, 21, 19, 15], [2, 14, 27, 6, 7], [22, 25, 18, 24, 9], [26, 21, 15, 18, 17], [7, 11, 22, 17, 21], [3, 11, 3, 20, 16], [22, 13, 18, 21, 11], [4, 27, 21, 20, 25], [4, 1, 13, 5, 13], [16, 19, 6, 6, 25], [19, 10, 14, 12, 14], [18, 13, 13, 16, 3], [12, 7, 26, 26, 12], [25, 21, 12, 23, 22], [11, 4, 24, 5, 27], [25, 10, 10, 26, 27], [8, 25, 20, 6, 23], [8, 0, 12, 26, 14], [7, 11, 6, 27, 26], [6, 24, 22, 23, 19]]
Pretty late, but for posterity: the easiest solution is to pick dimensions, then build up from the element strategy.
from hypothesis.strategies import composite, integers, lists
#composite
def complicated_rectangles(draw, max_N):
list_len = draw(integers(1, max_N))
sublist_len = draw(integers(0, 5))
element_strat = integers(0, min(list_len, 5))
sublist_strat = lists(
element_strat, min_size=sublist_len, max_size=sublist_len)
return draw(lists(
sublist_strat, min_size=list_len, max_size=list_len))

Efficient way to find index of elements in a large list of integers starting with max to min elements

I have a large list of integers unsorted, numbers might be duplicated. I would like to create another list which is a list of sub-lists of indexes from the first list starting with max element to min, in decreasing order.
For example, if I have a list like this:
list = [4, 1, 4, 8, 5, 13, 2, 4, 3, 7, 14, 4, 4, 9, 12, 1, 6, 14, 10, 8, 6, 4, 11, 1, 2, 11, 3, 9]
The output should be:
indexList = [[10, 17], [5], [14], [22, 25], [18], [13, 27], [3, 19], [9], [16, 20], [4], [0, 2, 7, 11, 12, 21], [8, 26], [6, 24], [1, 15, 23]]
where, [10, 17] is the index of where '14' is present and so on...
Shared my code below. Profiling it using cProfile for a list of around 9000 elements takes around ~6 seconds.
def indexList(list):
# List with sorted elements
sortedList = sorted(list, reverse = True)
seen = set()
uSortedList = [x for x in sortedList if x not in seen and not seen.add(x)]
indexList = []
for e in uSortedList:
indexList.append([i for i, j in enumerate(list) if j == e])
return indexList
Here you go:
def get_list_indices(ls):
indices = {}
for n, i in enumerate(ls):
try:
indices[i].append(n)
except KeyError:
indices[i] = [n]
return [i[1] for i in sorted(indices.items(), reverse=True)]
test_list = [4, 1, 4, 8, 5, 13, 2, 4, 3, 7, 14, 4, 4, 9, 12, 1, 6, 14, 10, 8, 6, 4, 11, 1, 2, 11, 3, 9]
print(get_list_indices(test_list))
Based on some very basic testing, it is about twice as fast as the code you posted.

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