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Recursively generating a list of lists in a triangular format given a height and value

I recently started looking into recursion to clean up my code and "up my game" as it were. As such, I'm trying to do things which could normally be accomplished rather simply with loops, etc., but practicing them with recursive algorithms instead.
Currently, I am attempting to generate a two-dimensional array which should theoretically resemble a sort of right-triangle in an NxN formation given some height n and the value which will get returned into the 2D-array.
As an example, say I call: my_function(3, 'a');, n = 3 and value = 'a'
My output returned should be: [['a'], ['a', 'a'], ['a', 'a', 'a']]
[['a'],
['a', 'a'],
['a', 'a', 'a']]
Wherein n determines both how many lists will be within the outermost list, as well as how many elements should successively appear within those inner-lists in ascending order.
As it stands, my code currently looks as follows:
def my_function(n, value):
base_val = [value]
if n == 0:
return [base_val]
else:
return [base_val] + [my_function(n-1, value)]
Unfortunately, using my above example n = 3 and value = 'a', this currently outputs: [['a'], [['a'], [['a'], [['a']]]]]
Now, this doesn't have to get formatted or printed the way I showed above in a literal right-triangle formation (that was just a visualization of what I want to accomplish).
I will answer any clarifying questions you need, of course!

return [base_val]
Okay, for n == 0 we get [[value]]. Solid. Er, sort of. That's the result with one row in it, right? So, our condition for the base case should be n == 1 instead.
Now, let's try the recursive case:
return [base_val] + [my_function(n-1, value)]
We had [[value]], and we want to end up with [[value], [value, value]]. Similarly, when we have [[value], [value, value]], we want to produce [[value], [value, value], [value, value, value]] from it. And so on.
The plan is that we get one row at the moment, and all the rest of the rows by recursing, yes?
Which rows will we get by recursing? Answer: the ones at the beginning, because those are the ones that still look like a triangle in isolation.
Therefore, which row do we produce locally? Answer: the one at the end.
Therefore, how do we order the results? Answer: we need to get the result from the recursive call, and add a row to the end of it.
Do we need to wrap the result of the recursive call? Answer: No. It is already a list of lists. We're just going to add one more list to the end of it.
How do we produce the last row? Answer: we need to repeat the value, n times, in a list. Well, that's easy enough.
Do we need to wrap the local row? Answer: Yes, because we want to append it as a single item to the recursive result - not concatenate all its elements.
Okay, let's re-examine the base case. Can we properly handle n == 0? Yes, and it makes perfect sense as a request, so we should handle it. What does our triangle look like with no rows in it? Well, it's still a list of rows, but it doesn't have any rows in it. So that's just []. And we can still append the first row to that, and proceed recursively. Great.
Let's put it all together:
if n == 0:
return []
else:
return my_function(n-1, value) + [[value] * n]
Looks like base_val isn't really useful any more. Oh well.
We can condense that a little further, with a ternary expression:
return [] if n == 0 else (my_function(n-1, value) + [[value] * n])

You have a couple logic errors: off-by-1 with n, growing the wrong side (critically, the non-base implementation should not use a base-sized array), growing by an array of the wrong size. A fixed version:
#!/usr/bin/env python3
def my_function(n, value):
if n <= 0:
return []
return my_function(n-1, value) + [[value]*n]
def main():
print(my_function(3, 'a'))
if __name__ == '__main__':
main()
Since you're returning mutable, you can get some more efficiency by using .append rather than +, which would make it no longer functional. Also note that the inner mutable objects don't get copied (but since the recursion is internal this doesn't really matter in this case).
It would be possible to write a tail-recursive version of this instead, by adding a parameter.
But python is a weird language for using unnecessary recursion.

The easiest way for me to think about recursive algorithms is in terms of the base case and how to build on that.
The base case (case where no recursion is necessary) is when n = 1 (or n = 0, but I'm going to ignore that case). A 1x1 "triangle" is just a 1x1 list: [[a]].
So how do we build on that? Well, if n = 2, we can assume we already have that base case value (from calling f(1)) of [[a]]. So we need to add [a, a] to that list.
We can generalize this as:
f(1) = [[a]]
f(n > 1) = f(n - 1) + [[a] * n]
, or, in Python:
def my_function(n, value):
if n == 1:
return [[value]]
else:
return my_function(n - 1, value) + [[value] * n]

While the other answers proposed another algorithm for solving your Problem, it could have been solved by correcting your solution:
Using a helper function such as:
def indent(x, lst):
new_lst = []
for val in lst:
new_lst += [x] + val
return new_lst
You can implement the return in the original function as:
return [base_val] + indent(value, [my_function(n-1, value)])
The other solutions are more elegant though so feel free to accept them.
Here is an image explaining this solution.
The red part is your current function call and the green one the previous function call.
As you can see, we also need to add the yellow part in order to complete the triangle.
These are the other solutions.
In these solutions you only need to add a new row, so that it's more elegant overall.

Why do slicing and sorted fail to reverse list in place?

I am doing the NextPermutation problem on LeetCodeN(https://leetcode.com/problems/next-permutation/). The problem needs me to sort in-place. Here's the problem (Source: LeetCode):
"Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).
The replacement must be in-place and use only constant extra memory.
Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1
"[the end]
Tried nums = nums[::-1] and nums = sorted(nums) and both failed the test case nums = [3,2,1] because the array appears to be unsorted. I also printed nums after these assignments and stdout is [1,2,3], which is correct. This error does not make sense because the last line of code nums[inverse_pt+1:] = sorted(nums[inverse_pt+1:]) modified my list as intended. Lastly, I tried nums.sort(), which sorts in-place, and this worked out.
I've read threads on reverse list, slicing.
def nextPermutation(self, nums: List[int]) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
if not nums or len(nums) == 1:
return
# find inverse pt
# O(1) because only 1 variable declared
inverse_pt = len(nums) - 2
while inverse_pt >= 0 and nums[inverse_pt+1] <= nums[inverse_pt]:
inverse_pt -= 1
if inverse_pt < 0:
nums = sorted(nums)
print(nums) # [1,2,3]
return
# find a smallest elem that's bigger than inverse_pt
for i in range(len(nums)-1, inverse_pt, -1):
if nums[i] > nums[inverse_pt]:
nums[i], nums[inverse_pt] = nums[inverse_pt], nums[i]
break
# sort what's after new elem at inverse_pt
nums[inverse_pt+1:] = sorted(nums[inverse_pt+1:])

They don't sort in place because that's not how they're designed.
Slicing always creates a new list that has some subset of elements from the original list in some order.
The sorted() method doesn't sort the list, it returns a sorted version of the list:
sorted(iterable, /, *, key=None, reverse=False)
Return a new list containing all items from the iterable in ascending order.
^^^^^^^^^^
Same with the reversed() method.
If you want to sort a list in-place, just do lst.sort():
sort(self, /, *, key=None, reverse=False)
Stable sort *IN PLACE*.
In general, this plays into part of python's design philosophy. In general, a method will either mutate its object or return a new, mutated version of the object without changing it. There are a couple of notable exceptions, mostly list.pop() (which is important for data science reasons and can be considered as a shorthand for two sequential operations), but slicing and sorted() (and reversed()) follow this paradigm.

sub-sum from a list without loops

So i'm studying recursion and have to write some codes using no loops
For a part of my code I want to check if I can sum up a subset of a list to a specific number, and if so return the indexes of those numbers on the list.
For example, if the list is [5,40,20,20,20] and i send it with the number 60, i want my output to be [1,2] since 40+20=60.
In case I can't get to the number, the output should be an empty list.
I started with
def find_sum(num,lst,i,sub_lst_sum,index_lst):
if num == sub_lst_sum:
return index_lst
if i == len(sum): ## finished going over the list without getting to the sum
return []
if sub_lst_sum+lst[i] > num:
return find_sum(num,lst,i+1,sub_lst_sum,index_lst)
return ?..
index_lst = find_sum(60,[5,40,20,20,20],0,0,[])
num is the number i want to sum up to,
lst is the list of numbers
the last return should go over both the option that I count the current number in the list and not counting it.. (otherwise in the example it will take the five and there will be no solution).
I'm not sure how to do this..

Here's a hint. Perhaps the simplest way to go about it is to consider the following inductive reasoning to guide your recursion.
If
index_list = find_sum(num,lst,i+1)
Then
index_list = find_sum(num,lst,i)
That is, if a list of indices can be use to construct a sum num using elements from position i+1 onwards, then it is also a solution when using elements from position i onwards. That much should be clear. The second piece of inductive reasoning is,
If
index_list = find_sum(num-lst[i],lst,i+1)
Then
[i]+index_list = find_sum(num,lst,i)
That is, if a list of indices can be used to return a sum num-lst[i] using elements from position i+1 onwards, then you can use it to build a list of indices whose respective elements sum is num by appending i.
These two bits of inductive reasoning can be translated into two recursive calls to solve the problem. Also the first one I wrote should be used for the second recursive call and not the first (question: why?).
Also you might want to rethink using empty list for the base case where there is no solution. That can work, but your returning as a solution a list that is not a solution. In python I think None would be a the standard idiomatic choice (but you might want to double check that with someone more well-versed in python than me).
Fill in the blanks
def find_sum(num,lst,i):
if num == 0 :
return []
elif i == len(lst) :
return None
else :
ixs = find_sum(???,lst,i+1)
if ixs != None :
return ???
else :
return find_sum(???,lst,i+1)

Extracting the subsequence of maximum length from a sequence [PYTHON] [duplicate]

This question already has an answer here:
Longest increasing unique subsequence
(1 answer)
Closed 6 years ago.
I have a sequence of values [1,2,3,4,1,5,1,6,7], and I have to find the longest subsequence of increasing length. However, the function needs to stop counting once it reaches a number lower than the previous one. The answer in this sequence in that case is [1,2,3,4]. As it has 4 values before being reset. How would I write the Python code for this?
Note: Finding the "longest increasing subsequence" seems to be a common challenge and so searching online I find a lot of solutions that would count for the entire length of the sequence, and return a subsequence of increasing values, ignoring any decrease, so in this case it would return [1,2,3,4,5,6,7]. That is not what I'm looking for.
It needs to count each subsequence, and reset the count upon reaching a number lower than the previous one. It then needs to compare all the subsequences counted, and return the longest one.
Thanks in advance.

Consider a function that generates all possible ascending subsequences, you would start with an empty list, add items until one element was less (or equal to?) the the previous at which point you save (yield) the subsequence and restart with a new subsequence.
One implementation using a generator could be this:
def all_ascending_subsequences(sequence):
#use an iterator so we can pull out the first element without slicing
seq = iter(sequence)
try: #NOTE 1
last = next(seq) # grab the first element from the sequence
except StopIteration: # or if there are none just return
#yield [] #NOTE 2
return
sub = [last]
for value in seq:
if value > last: #or check if their difference is exactly 1 etc.
sub.append(value)
else: #end of the subsequence, yield it and reset sub
yield sub
sub = [value]
last = value
#after the loop we send the final subsequence
yield sub
two notes about the handling of empty sequences:
To finish a generator a StopIteration needs to be
raised so we could just let the one from next(seq) propegate out - however when from __future__ import generator_stop is in
effect it would cause a RuntimeError so to be future compatible we
need to catch it and explicitly return.
As I've written it passing an empty list to
all_ascending_subsequences would generate no values, which may not
be the desired behaviour. Feel free to uncomment the yield [] to
generate an empty list when passed an empty list.
Then you can just get the longest by calling max on the result with key=len
b = [1,2,3,4,1,5,1,6,7]
result = max(all_ascending_subsequences(b),key=len)
print("longest is", result)
#print(*all_ascending_subsequences(b))

b = [4,1,6,3,4,5,6,7,3,9,1,0]
def findsub(a):
start = -1
count = 0
cur = a[0]
for i, n in enumerate(a):
if n is cur+1:
if start is -1:
start = i - 2
count=1
count+=1
cur = n
if n < cur and count > 1:
return [a[j] for j in range(start,start+count+1)]
print findsub(b)
A somewhat sloppy algorithm, but I believe it does what you want. Usually i would not have simply shown you code, but I suspect that is what you wanted, and I hope you can learn from it, and create your own from what you learn.
a slightly better looking way because I didn't like that:
b = [1,2,0,1,2,3,4,5]
def findsub(a):
answer = [a[0]]
for i in a[1:]:
if answer[-1] + 1 is i:
answer.append(i)
if not len(answer) > 1:
answer = [i]
elif i < answer[-1] and len(answer) > 1:
return answer
return answer
print findsub(b)

You need to do the following:
Create a function W that given a list, returns the index of the last item which is not strictly increasing from the start of the list.
For example, given the following lists: [1,2,3,4,1,2,3], [4,2,1], [5,6,7,8], it should return 4, 1, 4, respectively for each list
Create a variable maxim and set the value to 0
Repeatedly do the following until your list is empty
Call your function W on your list, and let's call the return value x
if x is greater than maxim
set maxim to x
At this point if you wish to store this sequence, you can use the list-slice notation to get that portion of your list which contains the sequence.
delete that portion of your list from the list
Finally, print maxim and if you were storing the parts of your list containing the longest sequence, then print the last one you got

What is the purpose of the fourth line of code particularly "l[j][-1]" here? [duplicate]

This question already has answers here:
Understanding slicing
(38 answers)
Closed 7 years ago.
The ff is a code to find the longest increasing subsequence of an array/list in this case it is d. The 4th line confuses me like what does l[j][-1] mean. Let's say i=1. What would l[j][-1] be?
d = [10,300,40,43,2,69]
l = []
for i in range(len(d)):
l.append(max([l[j] for j in range(i) if l[j][-1] < d[i]] or [[]], key=len) + [d[i]])

When I hit something like this, I consider that either I'm facing a smart algorithm that's beyond my trivial understanding, or that I'm just seeing tight code written by somebody while their mind was well soaked with the problem at hand, and not thinking about the guy who'd try to read it later (that guy most often being the writer himself, six months later... unless he's an APL hacker...).
So I take the code and try deobfuscate it. I copy/paste it into my text editor, and from there I rework it to split complex statements or expressions into smaller ones, and to assign intermediate values to variables with meaningful names. I do so iteratively, peeling off one layer of convolution at a time, until the result makes sense to me.
I did just that for your code. I believe you'll understand it easily by yourself in this shape. And me not being all too familiar with Python, I commented a few things for my own understanding at least as much as for yours.
This resulting code could use a bit of reordering and simplification, but I left in the form that would map most easily to the original four-liner, being the direct result of its expansion. Here we go :
data = [10,300,40,43,2,69]
increasingSubsequences = []
for i in range(len(data)):
# increasingSubsequences is a list of lists
# increasingSubsequences[j][-1] is the last element of increasingSubsequences[j]
candidatesForMoreIncrease = []
for j in range(i):
if data[i] > increasingSubsequences[j][-1]:
candidatesForMoreIncrease.append( increasingSubsequences[j] )
if len(candidatesForMoreIncrease) != 0:
nonEmpty = candidatesForMoreIncrease
else:
nonEmpty = [[]]
longestCandidate = max( nonEmpty, key=len ) # keep the LONGEST of the retained lists
# with the empty list included as a bottom element
# (stuff + [data[i]]) is like copying stuff and and then appending data[i] to the copy
increasingSubsequences.append( longestCandidate + [data[i]] )
print "All increasingSubsequences : ", increasingSubsequences
print "The result you expected : ", max(increasingSubsequences, key=len)

This looks like python to me.
max(arg1, arg2, *args[, key])
Returns the largest item in an iterable or the largest of two or more arguments.
If one positional argument is provided, iterable must be a non-empty iterable (such as a non-empty string, tuple or list). The largest item in the iterable is returned. If two or more positional arguments are provided, the largest of the positional arguments is returned.
The optional key argument specifies a one-argument ordering function like that used for list.sort(). The key argument, if supplied, must be in keyword form (for example, max(a,b,c,key=func)).
In this case it appears the variable l = [] is going to be appended with the largest iterable or the largest of 2 arguments from the stuff inside of the max function.
iterable 1:
[l[j] for j in range(i) if l[j][-1] < d[i]]
or
[[]]
Sometimes in words it helps too:
j in l for j in range i if j in l times -1 is less than i in d or [[]]
iterable 2:
key=len
Add [d[i]] to the one that returns the max result
Append this result to l = [] making l = [result]
Append simply adds an item to the end of the list; equivalent to a[len(a):] = [x].