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Working on some example questions, the particular one asks to make a function which would take a list and return a new one which would make every ascending sublist in the list go in descending order and leave the descending sublists as they are. For example, given the list [1,2,3,4,5], I need the list [5,4,3,2,1] or given a list like [1,2,3,5,4,6,7,9,8] would return [5,3,2,1,9,7,6,4,8]
Here's what I have so far, but it does not do anything close to what I'd like it to do:
def example3(items):
sublst = list()
for i in items:
current_element = [i]
next_element = [i+1]
if next_element > current_element:
sublst = items.reverse()
else:
return items
return sublst
print (example3([1,2,3,2])) #[[1, 2, 3, 2], [1, 2, 3, 2], [1, 2, 3, 2], [1, 2, 3, 2]]
EDIT:
I feel like people are a little confused as to what I want to do in this case, heres a better example of what I'd like my function to do. Given a list like: [5, 7, 10, 4, 2, 7, 8, 1, 3] I would like it to return [10, 7, 5, 4, 8, 7, 2, 3, 1]. As you can see all the sublists that are in descending order such as ([5,7,10]) gets reversed to [10, 7, 5].
It was a bit challenging to figure out what you need.
I think you want something like as follows:
import random
l = [5, 7, 10, 4, 2, 7, 8, 1, 3]
bl =[]
while True:
if len(l) == 0:
break
r = random.randint(0, len(l))
bl.extend(l[r:None:-1])
l = l[r+1:]
print(bl)
Out1:
[10, 7, 5, 4, 8, 7, 2, 3, 1]
Out2:
[10, 7, 5, 2, 4, 1, 8, 7, 3]
Out3:
[3, 1, 8, 7, 2, 4, 10, 7, 5]
Out4:
[2, 4, 10, 7, 5, 3, 1, 8, 7]
etc.
If you want a specific reverse random list:
import random
loop_number = 0
while True:
l = [5, 7, 10, 4, 2, 7, 8, 1, 3]
bl =[]
while True:
if len(l) == 0:
break
r = random.randint(0, len(l))
bl.extend(l[r:None:-1])
l = l[r+1:]
loop_number += 1
if bl == [10, 7, 5, 4, 8, 7, 2, 3, 1]:
print(bl)
print("I tried {} times".format(loop_number))
break
Out:
[10, 7, 5, 4, 8, 7, 2, 3, 1]
I tried 336 times
The general algorithm is to keep track of the current ascending sublist you are processing using 2 pointers, perhaps a "start" and "curr" pointer. curr iterates over each element of the list. As long as the current element is greater than the previous element, you have an ascending sublist, and you move curr to the next number. If the curr number is less than the previous number, you know your ascending sublist has ended, so you collect all numbers from start to curr - 1 (because array[curr] is less than array[curr - 1] so it can't be part of the ascending sublist) and reverse them. You then set start = curr before incrementing curr.
You will have to deal with the details of the most efficient way of reversing them, as well as the edge cases with the pointers like what should the initial value of start be, as well as how to deal with the case that the current ascending sublist extends past the end of the array. But the above paragraph should be sufficient in getting you to think in the right direction.
Let's say I have a list
A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
I want to increase every 3rd number by a value of 5 to result in
A = [1, 2, 8, 4, 5, 11, 7, 8, 14, 10]
My gut tells me something along the lines of
A[::3] = [x + 5 for x in A]
OR using the loop below with replace somehow integrated
for num in range(0, len(A), 3):
A = num + 5
Send help...thanks in advance.
You almost had it with your first attempt.
Modify the slice to start at the 3rd element (index 2) per your example, and make sure to read the same slice that you're writing to:
A[2::3] = [x+5 for x in A[2::3]]
i think this will do it
A=[x+5 if i%3==0 else x for i,x in enumerate(A,1)]
You could make a function for it which would account for duplicate values or unsorted values. If my assumptions are not correct, you could easily adjust the function to get your values correct.
A = [1, 2, 3, 4, 5, 6, 10, 8, 9, 7]
def list_modifier(passed_list):
passed_list.sort()
mod_list = list(set(passed_list))
out_list = {i:i if (mod_list.index(i)+1)%3 != 0 else (i+5) for i in mod_list }
passed_list = [out_list[i] for i in out_list]
return passed_list
list_modifier(A)
Returned list looks like this:
[1, 2, 8, 4, 5, 11, 7, 8, 14, 10]
I have a dictionaries inside a list like this:
sample_dict = [{1: [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], \
[1, 2, 3, 4, 5], \
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]]}, \
{2: [[3, 4, 6, 7, 8, 9, 10, 11], [1, 2, 3, 6, 10], []]}]
Now, I would like to check the key 1's first value in the list with key 2's first value.
something like this,
Compare Values (first value of list of lists of key 1)
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
with (first value of list of lists of key 2)
[3, 4, 6, 7, 8, 9, 10, 11]
If they are a match I would like to append it to a new list matching_list, if not I would to append non-matching values into another list non_matching_list.
This is what I tried so far,
matching_list = []
non_matching_list = []
for each_dict in sample_dict:
current_dict_values = []
for key, value_list in each_dict.items():
temp_dict_values = []
for value in value_list:
temp_dict_values.append(value)
.... don't know how to keep track of key 1's first list of lists values.
I was thinking of creating a temporary list to keep track of key 1 list values, but I am stuck and not sure how to proceed.
My final output should be like this:
matching_list = [[3,4,6,7,8,9,10], [1,2,3], []]
non_matching_list = [[1,2,5,11],[4,5,6,10],[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]]
How can I achieve my output? Any ideas would be great.
This can be achieved by converting lists to sets to make operations like symmetric_difference() and intersection() for your non_matching_list and matching_list respectively.
Here is one of the solutions:
matching_list, non_matching_list = [], []
for lists1, lists2 in zip(sample_dict[0].values(), sample_dict[1].values()):
for l1, l2 in zip(lists1, lists2):
matching_list.append(list(set(l1) & set(l2)))
non_matching_list.append(list(set(l1).symmetric_difference(set(l2))))
Note that using set(l1) & set(l2) is same as set(l1).intersection(set(l2)), so basically it's an intersection operation here.
I'm also using builtin zip() function to aggregate elements from each of the iterables ( both lists ).
Have a list arr = [1,3,4,5,2,3,4,2,5,7,3,8,1,9,6,2,1,2,1,3,4,3,4,6,9]
want to remove the duplicate values so that the original list should contains single instances of all elements. Do not want to create a extra list and append the elements from list. Also do not want to use inbuilt "set".
Tried to do that with some code as below:
l = len(arr)
for x in range(l):
for y in range(x+1,l):
if arr[x] == arr[y]:
del arr[y]
Tried the above code and its throwing error
"IndexError: list index out of range"
What I understand is whiling deleting the value the size of the list is changing for which its throwing the error. So I made the below changes. But still its failing with same error:
l = len(arr)
for x in range(l):
for y in range(x+1,l):
if arr[x] == arr[y]:
t = y
del arr[y]
y = t - 1
Can some one help me out on this?
Thanks in Advance.
You are trying to make the code more efficient by caching the length of the list in the local variable l. However, that is not helpful because the list is being trimmed inside the loop, and you are not keeping the cached length variable in sync.
for index in range(len(arr)-1,0,-1):
if arr[index] in arr[:index]:
del arr[index]
By going backwards through the array and looking for earlier occurrences of each element, you can avoid having to worry about the length of the list changing all the time.
This method also preserves the order in which elements occur in the original array. Note the instruction is to only remove duplicates (a.k.a. subsequent occurrences).
For example the list [9,3,4,3,5] should reduce to [9,3,4, 5] as the second occurrence of 3 is considered a duplicate and should be removed.
How about this approach:
>>> set(arr)
set([1, 2, 3, 4, 5, 6, 7, 8, 9]) #Just to compare it with the results below.
>>> arr = [1,3,4,5,2,3,4,2,5,7,3,8,1,9,6,2,1,2,1,3,4,3,4,6,9]
>>> arr.sort()
>>> arr
[1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 7, 8, 9, 9]
>>> for i in arr:
while arr.count(i) > 1:
del arr[i]
>>> arr
[1, 2, 3, 4, 5, 6, 7, 8, 9]
Another approach is to find, after sorting your list, the length of the sublist to delete for each number:
>>> arr = [1,3,4,5,2,3,4,2,5,7,3,8,1,9,6,2,1,2,1,3,4,3,4,6,9]
>>> arr.sort()
>>> arr
[1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 7, 8, 9, 9]
>>> for i,j in enumerate(arr):
del arr[i+1:i+arr.count(j)]
>>> arr
[1, 2, 3, 4, 5, 6, 7, 8, 9]
I have a question about python. I have to sort a list of random numbers in a particular way (it's not allowed to use sort()). I'll try to explain:
I have to search for the smallest number, and swap this number with the number at the first position in the list.
Then, I search again for the smallest number, but this time ignore the first number in my list because this one is already sorted. So, I should start searching for the smallest number from the second number (index 1) till the end of the list. The smallest number then found, should be swapped with the second number in the list(so the index 1).
I hope you understand my problem. This is the code I wrote so far, but I get errors and/or the sorting isn't correct.
array = random_integers(10,size=10)
my_list = list(array)
for i in range(len(my_list)):
print my_list
a = min(my_list[i:len(my_list)])
b = my_list.index(a)
my_list[i],my_list[b]=my_list[b],my_list[i]
print my_list
I think there's a problem in my range, and a problem with the
a = min(my_list[i:len(my_list)])
I want to search for the smallest number, but not in the ENTIRE list how can I do this?
The problem occurs on this line:
b = my_list.index(a)
since this searches for the first occurrence of a in all of my_list. If the same number occurs twice, then b will always correspond to the smallest such index, which might be less than i. So you might end up moving a number which has already been sorted.
The obvious thing to try is to slice my_list before calling index:
my_list[i:].index(a)
but note that index will return values between 0 and N-i. We want numbers between i and N. So be sure to add i to the result:
b = my_list[i:].index(a)+i
Thus, the easiest way to fix your code as it presently exists is:
for i in range(len(my_list)):
a = min(my_list[i:])
b = my_list[i:].index(a)+i
my_list[i], my_list[b] = my_list[b], my_list[i]
but notice that min is searching through all the items in my_list[i:] and then the call to index is traversing the same list a second time. You could find b in one traversal like this:
b = min(range(i, N), key=my_list.__getitem__)
Demo:
import numpy as np
array = np.random.random_integers(10,size=10)
my_list = list(array)
N = len(my_list)
for i in range(N):
b = min(range(i, N), key=my_list.__getitem__)
my_list[i], my_list[b] = my_list[b], my_list[i]
print my_list
yields
[3, 10, 9, 6, 5, 3, 6, 8, 8, 4]
[3, 3, 9, 6, 5, 10, 6, 8, 8, 4]
[3, 3, 4, 6, 5, 10, 6, 8, 8, 9]
[3, 3, 4, 5, 6, 10, 6, 8, 8, 9]
[3, 3, 4, 5, 6, 10, 6, 8, 8, 9]
[3, 3, 4, 5, 6, 6, 10, 8, 8, 9]
[3, 3, 4, 5, 6, 6, 8, 10, 8, 9]
[3, 3, 4, 5, 6, 6, 8, 8, 10, 9]
[3, 3, 4, 5, 6, 6, 8, 8, 9, 10]
[3, 3, 4, 5, 6, 6, 8, 8, 9, 10]
If you want the smallest number from a list you can use min(). If you want a part of a list you can use list slicing: my_list[1:]. Put the two together and you get the smallest number from a part of your list. However, you don't need to do this, as you can .pop() from the list instead.
sorted_list = []
while my_list:
n = min(my_list)
sorted_list.append(my_list.pop(my_list.index(n)))
If you're using numpy arrays then instead of my_list.index(min(my_list)) you can use the .argmin() method.
While this type of sorting is good for an introduction, it is not very efficient. You may want to consider looking at the merge sort, and also the Python's built-in timsort.