def my_sort(array):
length_of_array = range(1, len(array))
for i in length_of_array:
value = array[i]
last_value = array[i-1]
if value<last_value:
array[i]=last_value
array[i-1]=value
my_sort(array)
return array
I know what the function does in general. Its a sorting alogarithm.... But i dont know how what each individual part/section does.
Well, I have to say that the best way to understand this is to experiment with it, learn what it is using, and, basically, learn Python. :)
However, I'll go through the lines one-by-one to help:
Define a function named my_sort that accepts one argument named array. The rest of the lines are contained in this function.
Create a range of numbers using range that spans from 1 inclusive to the length of array non-inclusive. Then, assign this range to the variable length_of_array.
Start a for-loop that iterates through the range defined in the preceding line. Furthermore, assign each number returned to the variable i. This for-loop encloses lines 4 through 9.
Create a variable value that is equal to the item returned by indexing array at position i.
Create a variable last_value that is equal to the item returned by indexing array at position i-1.
Test if value is less than last_value. If so, run lines 7 through 9.
Make the i index of array equal last_value.
Make the i-1 index of array equal value.
Rerun my_sort recursively, passing in the argument array.
Return array for this iteration of the recursive function.
When array is finally sorted, the recursion will end and you will be left with array all nice and sorted.
I hope this shed some light on the subject!
I'll see what I can do for you. The code, for reference:
def my_sort(array):
length_of_array = range(1, len(array))
for i in length_of_array:
value = array[i]
last_value = array[i-1]
if value<last_value:
array[i]=last_value
array[i-1]=value
my_sort(array)
return array
def my_sort(array):
A function that takes an array as an argument.
length_of_array = range(1, len(array))
We set the variable length_of_array to a range of numbers that we can iterate over, based on the number of items in array. I assume you know what range does, but if you don't, in short you can iterate over it in the same way you'd iterate over a list. (You could also use xrange() here.)
for i in length_of_array:
value = array[i]
last_value = array[-1]
What we're doing is using the range to indirectly traverse the array because there's the same total of items in each. If we look closely, though, value uses the i as its index, which starts off at 1, so value is actually array[1], and last_value is array[1-1] or array[0].
if value<last_value:
array[i]=last_value
array[i-1]=value
So now we're comparing the values. Let's say we passed in [3, 1, 3, 2, 6, 4]. We're at the first iteration of the loop, so we're essentially saying, if array[1], which is 1, is less than array[0], which is 3, swap them. Of course 1 is less than 3, so swap them we do. But since the code can only compare each item to the previous item, there's no guarantee that array will be properly sorted from lowest to highest. Each iteration could unswap a properly swapped item if the item following it is larger (e.g. [2,5,6,4] will remain the same on the first two iterations -- they will be skipped over by the if test -- but when it hits the third, 6 will swap with 4, which is still wrong). In fact, if we were to finish this out without the call to my_sort(array) directly below it, our original array would evaluate to [1, 3, 2, 3, 4, 6]. Not quite right.
my_sort(array)
So we call my_sort() recursively. What we're basically saying is, if on the first iteration something is wrong, correct it, then pass the new array back to my_sort(). This sounds weird at first, but it works. If the if test was never satisfied at all, that would mean each item in our original list was smaller than the next, which is another way (the computer's way, really) of saying it was sorted in ascending order to begin with. That's the key. So if any list item is smaller than the preceding item, we jerk it one index left. But we don't really know if that's correct -- maybe it needs to go further still. So we have to go back to the beginning and (i.e., call my_sort() again on our newly-minted list), and recheck to see if we should pull it left again. If we can't, the if test fails (each item is smaller than the next) until it hits the next error. On each iteration, this teases the same smaller number leftward by one index until it's in its correct position. This sounds more confusing than it is, so let's just look at the output for each iteration:
[3, 1, 3, 2, 6, 4]
[1, 3, 3, 2, 6, 4]
[1, 3, 2, 3, 6, 4]
[1, 2, 3, 3, 6, 4]
[1, 2, 3, 3, 4, 6]
Are you seeing what's going on? How about if we only look at what's changing on each iteration:
[3, 1, ... # Wrong; swap. Further work ceases; recur (return to beginning with a fresh call to my_sort()).
[1, 3, 3, 2, ... # Wrong; swap. Further work ceases; recur
[1, 3, 2, ... # Wrong; swap. Further work ceases; recur
[1, 2, 3, 3, 6, 4 # Wrong; swap. Further work ceases; recur
[1, 2, 3, 3, 4, 6] # All numbers all smaller than following number; correct.
This allows the function to call itself as many times as it needs to pull a number from the back to the front. Again, each time it's called, it focuses on the first wrong instance, pulling it one left until it puts it in its proper position. Hope that helps! Let me know if you're still having trouble.
Related
I want to create a funciton getCombinations that given a list of positive integers and a maximum amount, appends all the possible combinations of the given integers that sum to the given amount to a list outside of the function.
For example:
combinations=[]
getCombinations([5,2,1], 10)
print(combinations)
should return:
[[5, 5], [5, 2, 2, 1], [5, 2, 1, 1, 1], [5, 1, 1, 1, 1, 1], [2, 2, 2, 2, 2], ... , [1, 1, 1, 1, 1, 1, 1, 1, 1, 1,]
I tried to use a recursive function that loops over the given integers, and appends them to the list current_combinations.
If the sum of that list equals the given amount, it should append the current_combination.
If the sum is smaller it goes one level deeper and appends new numbers from the list.
combinations = []
def getCombinations(num_types, max_amount, current_combi=None):
if current_combi is None:
current_combi = []
for num in num_types:
current_combi.append(num)
if sum(current_combi) == max_amount:
combinations.append(current_combi)
elif sum(current_combi) < max_amount:
getCombinations(num_types, max_amount, current_combi)
current_combi = current_combi[:-1]
getCombinations([5, 2, 1], 10)
print(combinations)
But this only outputs a fraction of the anticipated result:
[[5, 5], [5, 2, 2, 1], [5, 2, 2, 1], [5, 2, 1, 2], [5, 2, 1, 1, 1], [5, 2, 2, 1], [5, 2, 2, 1], [5, 2, 1, 2], [5, 2, 1, 1, 1]]
Help would be much appreciated, thank you.
There are two main problems here.
First, there is the algorithmic problem, which is that in every recursive call, you start at the beginning of the list of possible values. That will inevitably lead to duplicated lists. You want the produced lists to be sorted in descending order (or, at least, in the same order as the original list), and you need to maintain that order by not recursing over values you have already finished with.
The second problem is subtler; it has to do with the way you handle the current_combi arrays, with emphasis on the plural. You shouldn't use multiple arrays; when you do, it's very easy to get confused. What you should do is use exactly one array, and only make a copy when you need to add it to the result.
You might need to pull out a pad of paper and a pencil and play computer to see what's going on, but I'll try to describe it. The key is that:
current_combi.append(num) modifies the contents of current_combi;
passing current_combi as a function argument does not create a new list;
current_combi = current_combi[:-1] does create a new list, and does not modify the old one.
So, you enter getCombinations and create a current_combi list. Then you push the first value onto the new list (so it's now [5] and recursively call getCombinations, passing the same list. Inside the recursive call, the first value is appended again onto the list (which is still the same list); that list is now [5, 5]. That's a valid result, so you add it to the accumulated results, and then you create a new current_combi. At this point, the original current_combi is still [5, 5] and the new one is [5]. Then the for loop continues (in the recursive call), but the rest of that loop no longer has access to the original current_combi. So we can fast forward to the end of the for loop, ignoring the recursive subcalls, and return to the top level.
When we return to the top level, current_combi is the list which was originally created, and that list had 5 appended to it twice, once in the top-level for loop and again when the first recursive call started. So it's still [5. 5], which is unexpected. A fundamental property of recursive backtracking is that the problem state variable be the same before and after each recursive call. But that property has been violated. So now at the end of the top-level for loop, an attempt is made to remove the 5 added at the beginning. But since that list is now [5, 5], removing the last element produces [5] instead of []. As a result, lists starting with 2 are never produced, and lists starting 5, 2 are produced twice.
OK, let's fix that. Instead of making copies of the list at uncontrolled points in the execution, we'll just use one list consistently, and make a copy when we add it to the accumulated results:
# This one still doesn't work. See below.
def getCombinations(num_types, max_amount, current_combi=None):
if current_combi is None:
current_combi = []
for num in num_types:
current_combi.append(num)
if sum(current_combi) == max_amount:
combinations.append(current_combi[:]) # Add a copy to results
elif sum(current_combi) < max_amount:
getCombinations(num_types, max_amount, current_combi)
current_combi.pop() # Restore current_combi
But that doesn't actually fix the first problem noted above: that the recursion should not reuse values which have already been used. Instead of looping over the values in num_types, we need to loop over its suffixes:
def getCombinations(num_types, max_amount, current_combi=None):
if current_combi is None:
current_combi = []
values = num_types[:] # Avoid modifying the caller's list
while values:
# values.pop(0) removes the first element in values and returns it
current_combi.append(values.pop(0))
if sum(current_combi) == max_amount:
combinations.append(current_combi[:]) # Add a copy to results
elif sum(current_combi) < max_amount:
getCombinations(values, max_amount, current_combi)
# Restore current_combi
current_combi.pop()
In the above, I was trying to roughly follow the logic of your original. However, the handling of values could be made more efficient by passing the starting index in the list instead of modifying the list. Also, there is no need to rescan the candidate combination in order to add up its value, since we can just add the value we just added (or, as in the following code, subtract it from the target). Finally, since a lot of the arguments to the recursive call are always the same, they can be replaced by a closure:
def getCombinations(num_types, max_amount):
results = []
candidate = []
def helper(first_index, amount_left):
for index in range(first_index, len(num_types)):
value = num_types[index]
if amount_left == value:
results.append(candidate + [value])
elif amount_left > value:
candidate.append(value)
helper(index, amount_left - value)
candidate.pop()
helper(0, max_amount)
return results
That's still not the optimal implementation, but I hope it shows how to evolve this implementation.
I'm trying to understand the logic here, why this code doesn't remove all elements in the list?
int_lst = [1, 2, 3, 4, 5, 6]
for i in int_lst:
int_lst.remove(i)
print(int_lst)
print(int_lst) #Output: [2, 4, 6]
When you remove an element, everything after that element effectively moves back one place. But the iterator doesn't know that: it just advance the pointer to the next element, skipping what should have been the next value bound to i.
In detail, i starts at 1. You remove it, and now 2 is in the current position of the iterator. The iterator advances, and now 3 is in the current position and bound to i.
In other words, every time you remove an element, you effectively advance the iterator, in addition to the explicit advancement the iterator will perform.
That happens because you are removing elements in the same list you are iterating through. You can easily fix it by avoiding this:
for i in list(int_lst):
int_lst.remove(i)
When running the code below, I can't understand why it doesn't remove all 1's? It keeps two of them and with my two print statements added before the if statement, it shows the program seemingly skipping over the last 2 of the 4 1's. I know of an easier way to do this, I'm just trying to understand the logic of why this isn't working.
numbers3 = [1, 1, 1, 1, 4, 5, 7, 7, 9]
print(numbers3.count(1))
for number in numbers3:
print(numbers3.count(number))
print(number)
if numbers3.count(number) > 1:
numbers3.remove(number)
print(numbers3)
I expect the out put to be [1, 4, 5, 7, 9]
But instead I get [1, 1, 4, 5, 7, 9]
Welcome to Stackoverflow.
The problem is that you are modifying the list you are iterating over, which is rarely a recipe for a happy result in Python (or most other languages).
It's fairly common to see questions from beginners that usefully tell us what the expected output is, but not why. Reading your code, I deduce (or rather guess) that you would like a list containing a single copy of each unique number.
One way to achieve your end is to build a separate list of the numbers you want. So this (untested) should work:
result = []
for number in numbers3:
if number not in result:
result.append(number)
A simpler way, which since Python 3.6 should retain ordering, would be
result = list(set(numbers3))
Thats because youre removing from the list meanwhile iterating over it. you remove [1,1] and then the 4 becomes the 2th index and it skips the other [1,1].
Suppose I have a list
a=[0,1,2,3,4,5,6,7,8,9]
I randomly choose an index using,
i = np.random.choice(np.arange(a.size))
and now, I want to select a symmetric interval around i of some size, say 2. So for example for the given list, if the index i = 5 is selected, I get something like
print(a[i-2:i+2+1])
>> [3, 4, 5, 6, 7]
This works fine
however, if i happens to be near one of the end points, i = 1. Using what I have i get,
print(a[i-2:i+2+1])
>> []
Instead I want something it to print an asymmetric interval, like [0, 1, 2, 3]
if i = 8
print(a[i-2:i+2+1])
>> [6, 7, 8, 9]
like I want it to too, so being near the end point doesn't seem to matter. The closest I have gotten to a solution is (say for i = 1)
print([a[0:i+3] if a[i-2:i+2+1] == [] else a[i-2:i+2+1] ])
>> [[0, 1, 2, 3]]
But this returns, [[0,1,2,3]] instead of [0,1,2,3]
Is there a nice way to do this in python/numpy using list comprehension or something else?
You just need go clip the lower index at zero:
>>> print(a[max(i-2,0):i+2+1])
[0, 1, 2, 3]
Without this, it can got into negative numbers. This has special meaning in slicing: negative indices count from the end of the list.
You've tripped across the right-end numbering of Python. You gave it the limits [-1:3], but -1 denotes the right-hand element. Since the "first" element is past the "last" element, the resulting slice is 0. You won't have this problem on the high indices, because there's no "wrap-around" on that end.
Simply drive the lower index to a rail of 0, using max.
print(a[max(i-2, 0)]:i+2+1])
The problem is that when you exit the bounds of the list, it returns an empty list. Try:
a=[0,1,2,3,4,5,6,7,8,9]
i = 1
interval=2
print( a[ max(i-2, 0) : min(i+2+1, len(a)) ] )
I just put max/min bounds on it so it doesn't escape it. Not very pythonic, but it's a quick fix.
I have a problem: i need to find an average of the list using this scheme:
First of all, we find an average of two elements, three elements..... len(list) elements and form a new list using averages. The use .pop() and find all averages again. Function should stop when len(list) == 2. Recursion should be used.
Example:
list: [-1, 4, 8, 1]
1 step:
find an average of [-1, 4], [-1, 4, 8], [-1, 4, 8, 1]
Then we form a new list: [1.5, 3.66..., 3] (averages)
Then find averages of new list: [1.5, 3.66...], [1.5, 3.66..., 3]
Then we form a new list: [2.5833.., 7.222...] (averages)
When len(list) == 2, find an average of this two elements.
Answer is 2.652777.
What should i write:
jada = []
while True:
print 'Lst elements:'
a = input()
if (a == ''):
break
jada.append(a)
print 'Lst is:' + str(Jada)
def keskmine(Jada):
for i in range(len(Jada) - 1):
...
jada.pop()
return keskmine(Jada)
Actually, this is a part of a homework, but i don't know how to solve it.
Accept the list as the function argument. If the list has one item, return that. Create two iterators from the list. Pop one item off one of the lists, zip them together, then find the averages of the zip results. Recurse.
In short, you're finding the "running average" from a list of numbers.
Using recursion would be helpful here. Return the only element when "len(lst) == 1" otherwise, compute the running average and recurse.
There are two parts in this assignment. First, you need to transform lists like [-1, 4, 8, 1] to lists like [1.5, 3.66, 3] (find the running averages). Second, you need to repeat this process with the result of the running averages until your list's length is 2 (or 1).
You can tackle the first problem (find the running averages) independently from the second. Finding the running average is simple, you first keep track of the running sum (e.g. if the list is [-1, 4, 8, 1] the running sum is [-1, 3, 11, 12]) and divide each elements by their respective running index (i.e. just [1, 2, 3, 4]), to get [-1/1, 3/2, 11/3, 12/4] = [-1, 1.5, 3.66, 3]. Then you can discard the first element to get [1.5, 3.66, 3].
The second problem can be easily solved using recursion. Recursion is just another form of looping, all recursive code can be transformed to a regular for/while-loops code and all looping code can be transformed to recursive code. However, some problems have a tendency towards a more "natural" solution in either recursion or looping. In my opinion, the second problem (repeating the process of taking running averages) is more naturally solved using recursion. Let's assume you have solved the first problem (of finding the running average) and we have a function runavg(lst) to solve the first problem. We want to write a function which repeatedly find the running average of lst, or return the average when the lst's length is 2.
First I'll give you an explanation, and then some pseudo code, which you'll have to rewrite in Python. The main idea is to have one function that calls itself passing a lesser problem with each iteration. In this case you would like to decrease the number of items by 1.
You can either make a new list with every call, or reuse the same one if you'd like. Before passing on the list to the next iteration, you will need to calculate the averages thus creating a shorter list.
The idea is that you sum the numbers in a parameter and divide by the number of items you've added so far into the appropriate index in the list. Once you are done, you can pop the last item out.
The code should look something like this: (indexes in sample are zero based)
average(list[])
if(list.length == 0) // Check input and handle errors
exit
if(list.length == 1) // Recursion should stop
return list[0] // The one item is it's own average!
// calculate the averages into the list in indices 0 to length - 2
list.pop() // remove the last value
return average(list) // the recursion happens here
This is also an opportunity to use python 3.x itertools.accumulate:
From docs:
>>> list(accumulate(8, 2, 50))
[8, 10, 60]
Then, you only need to divide each item by its index increased by 1, eliminate the first element and repeat until finished
For example, this works for any list of any length, doing most of the above-indicated steps inside a list comprehension:
>>> from itertools import accumulate
>>> a = [-1, 4, 8, 1]
>>> while len(a) > 1:
a = [item / (index + 1) for (index, item) in enumerate(accumulate(a)) if index > 0]
>>> print(a)
[2.6527777777777777]