I looked for answers but couldn't find anything.
>my_list = [7,8,9]
>my_list[2:1:-1]
[9]
>my_list[2:0:-1]
[9, 8]
>my_list[2:-1:-1]
[] # I would expect [9, 8, 7]
I am aware there are other ways to reverse a list, but I'm curious why the above doesn't work? my_list[2,None,-1] does work.
If you want to iterate from the back until the first element, leave the second element empty:
my_list[2::-1] well or write it sorter my_list[::-1]
Related
I am a newbie to the programming and trying to understand how things work.
I couldn't get my program to iterate through numpy.array normally, so I decided to try one level simple iteration through list. But still it wouldn't work!
The code is as follows:
my_list = [1, 2, 3, 4, 5, 6, 7]
for i in my_list:
print(my_list[i])
The output is:
So it doesn't take the my_list[0] index of some reason and comes out of range.
Could you please help me to understand why?
It's not clear what exactly you're trying to do. When you iterate over an iterable like a list with
for i in my_list:
each i is each member of the list, not the index of the member of the list. So, in your case, if you want to print each member of the list, use
for i in my_list:
print(i)
Think about it: what if the 3rd member of the list was 9, for example? Your code would be trying to print my_list[9], which doesn't exist.
As pointed out that's not how you should iterate over the elements of the list.
But if you persist, your loop be should over the range(my_list), at the moment you're indexing by the values of the list, and since the length is 7, the last valid index is 6, not 7.
You get an IndexError, because you are looping through the values, which means, that your first value for i is 1 and the last one is 7. Because 7 is an invalid Index for this list you get an IndexError. A suitable code would be:
my_list = [1, 2, 3, 4, 5, 6, 7]
for i in my_list:
print(i)
I am attempting to add items to a new list from a primary list, then remove those moved items from the primary list. Essentially use and discard.
I have already attempted to use list comprehensions (Remove all the elements that occur in one list from another).
Then I attempted to use a for loop as well as an if statement to check for the elements and remove them, though when I printed the original list once again, nothing changed.
Not sure what I am doing wrong but it is extremely frustrating:
your_hand_list = []
computer_hand_list = []
computer_hand_list.append (random.sample(card_list, 5))
your_hand_list.append (random.sample(card_list, 5))
print (your_hand_list, computer_hand_list)
for card in your_hand_list and computer_hand_list:
if card in card_list:
card_list.remove(card)
Edit: Adding your code made it much easier to understand your issue. Specifically, the problem is this statement:
for card in your_hand_list and computer_hand_list:
The Python and operator does not work the way you were expecting it to in this context. Instead, the Python standard itertools library has a function called chain that will solve your problem. Here is a (greatly) simplified version of your code for illustration purposes.
Example:
from itertools import chain
card_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
your_hand_list = [1, 4, 7]
computer_hand_list = [3, 9, 10]
for card in chain(your_hand_list, computer_hand_list):
if card in card_list:
card_list.remove(card)
print(card_list)
Output:
[2, 5, 6, 8]
I am trying to sort an unsorted list [4, 5, 9, 9, 0, 1, 8]
The list has two repeated elements. I have tried to approach the question by having a loop that goes through each element comparing each element with the next in the list and then placing the smaller element at the start of the list.
def sort(ls:
ls[x]
x = [4, 5, 9, 9, 0, 1, 8]
while len(x) > 0:
for i in the range(0, len(x)):
lowest = x[i]
ls.append(lowest)
Please, could someone explain where I am going wrong and how the code should work?
It may be that I have incorrectly thought about the problem and my reasoning for how the code should work is not right
I do not know, if this is exactly what you are looking for but try: sorted(ListObject).
sorted() returns the elements of the list from the smallest to the biggest. If one element is repeated, the repeated element is right after the original element. Hope that helped.
Yes, you can try x.sort() or sorted(x). Check this out https://www.programiz.com/python-programming/methods/built-in/sorted. Also, in your program I don't see you making any comparisons, for example, if x[i] <= x[i+1] then ...
This block of code is just gonna append all the elements in the same order, till n*n times.
Also check this https://en.wikipedia.org/wiki/Insertion_sort
For a built-in Python function to sort, let y be your original list, you can use either sorted(y) or y.sort().Keep in mind that sorted(y) will return a new list so you would need to assign it to a variable such as x=sorted(y); whereas if you use x.sort() it will mutate the original list in-place, so you would just call it as is.
If you're looking to actually implement a sorting function, you can try Merge Sort or Quick Sort which run in O (n log n) in which will handle elements with the same value. You can check this out if you want -> https://www.geeksforgeeks.org/python-program-for-merge-sort/ . For an easier to understand sorting algorithm, Insertion or Bubble sort also handle duplicate as well but have a longer runtime O (n^2) -> https://www.geeksforgeeks.org/python-program-for-bubble-sort/ .
But yea, I agree with Nameet, what you've currently posted looks like it would just append in the same order.
Try one of the above suggestions and hopefully this helps point you in the right direction to if you're looking for a built-in function or to implement a sort, which can be done in multiple ways with different adv and disadv to each one. Hope this helps and good luck!
There are several popular ways for sorting. take bubble sort as an example,
def bubbleSort(array):
x = len(array)
while(x > 1): # the code below make sense only there are at least 2 elements in the list
for i in range(x-1): # maximum of i is x-2, the last element in arr is arr[x-1]
if array[i] > array[i+1]:
array[i], array[i+1] = array[i+1], array[i]
x -= 1
return array
x = [4, 5, 9, 9, 0, 1, 8]
bubbleSort(x)
your code has the same logic as below
def sorts(x):
ls = []
while len(x) > 0:
lowest = min(x)
ls.append(lowest)
x.remove(lowest)
return ls
x = [4, 5, 9, 9, 0, 1, 8]
sorts(x)
#output is [0, 1, 4, 5, 8, 9, 9]
I need to be able to check if any items in one list are also in another list but in the same position. I have seen others but they return true or false. I need to know how many are in the same position.
So compare them directly!
This of course is assuming both lists are the same length:
a = [1, 1, 2, 3, 4, 5, 7, 8, 9]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9]
matches = 0
for index in range(len(a)):
if a[index] == b[index]:
matches += 1
print mat
Try it here!
overlap = set(enumerate(listA)).intersection(set(enumerate(listB))
print(len(overlap))
enumerate pairs up elements with their index, so you can check how many common element/ index pairs exist between the two lists.
One advantage of this approach (as opposed to iterating through either list yourself) is that it automatically takes care of the case where the lists have different lengths.
demo
Imagine a python list
a=[[1,2],[3,4],[5,6]]
how could I get a list out of this such that I get the elements of the list a that satisfy that the second element is bigger than 2?
In [32]: a=[[1,2],[3,4],[5,6]]
In [33]: [s for s in a if s[1]>2]
Out[33]: [[3, 4], [5, 6]]
So I guess the result you expect with your example is:
result=[[3,4], [5,6]]
In that case you'd just need to do:
result=[l for l in a if l[1]>2]
This is called list comprehension.