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How do I remove a character from an element in a list?
mylist = ['12:01', '12:02']
I want to remove the colon from the time stamps in a file, so I can more easily convert them to a 24hour time. Right now I am trying to loop over the elements in the list and search for the one's containing a colon and doing a substitute.
for num in mylist: re.sub(':', '', num)
But that doesn't seem to work.
The list comprehension solution is the most Pythonic one, but, there's an important twist:
mylist[:] = [s.replace(':', '') for s in mylist]
If you assign to
mylist, the barename, as in the other answer, rather than to
mylist[:], the "whole-list slice", as I recommend, you're really doing something very different than "replacing entries in the list": you're making a new list and just rebinding the barename that you were previously using to refer to the old list.
If that old list is being referred to by multiple names (including entries in containers), this rebinding doesn't affect any of those: for example, if you have a function which takes
mylist as an argument, the barename assignment has any effect only locally to the function, and doesn't alter what the caller sees as the list's contents.
Assigning to the whole-list slice,
mylist[:] = ..., alters the list object rather than mucking around with switching barenames' bindings -- now that list is truly altered and, no matter how it's referred to, the new value is what's seen. For example, if you have a function which takes
mylist as an argument, the whole-list slice assignment alters what the caller sees as the list's contents.
The key thing is knowing exactly what effect you're after -- most commonly you'll want to alter the list object, so, if one has to guess, whole-list slice assignment is usually the best guess to take;-). Performance-wise, it makes no difference either way (except that the barename assignment, if it keeps both old and new list objects around, will take up more memory for whatever lapse of time both objects are still around, of course).
Editing elements in a list in python, Looking to modify an item within a list in Python? If so, I'll show you the steps to accomplish this goal using a simple example. How to Modify Lists in Python. 1 Open a Python Shell window. You see the familiar Python prompt. 2 Type List1 =  and press Enter. 3 Type len(List1) and press Enter. 4 Type List1.append(1) and press Enter. 5 Type len(List1) and press Enter.
Use a list comprehension to generate a new list:
>>> mylist = ['12:01', '12:02'] >>> mylist = [s.replace(':', '') for s in mylist] >>> print mylist ['1201', '1202']
The reason that your solution doesn't work is that
re.sub returns a new string -- strings are immutable in Python, so
re.sub can't modify your existing strings.
How to Modify an Item Within a List in Python, Overview List is one of the simplest and most important data Home >> List Manipulation in Python. Apr. 07 Length - number of items in list In python 3 we use the index number of the list element to modify individual list elements. "Remember that python list starts with the index position 0 not 1. First element of the list has the index position 0, while the second element of the list has the index position 1. We use following syntax to modify a list item in python programming.
for i, num in enumerate(mylist): mylist[i] = num.replace(':','')
Python Lists and List Manipulation, Modify an element by using the index of the element; Remove an item from To update the required list element, you have to use the Python for loop. In the loop, you have to use the if condition to compare and change only the required element. Check the below-given example and change the required element as per your requirement.
You have to insert the return of re.sub back in the list. Below is for a new list. But you can do that for mylist as well.
mylist = ['12:01', '12:02'] tolist =  for num in mylist: a = re.sub(':', '', num) tolist.append(a) print tolist
Lists, mutability, and in-place methods, So it is permitted and easy to change elements in a list. Let's make a list: lst = [ 'alpha', 'beta', 'delta']. Let's replace the third element: lst[ In method 1a, Python doesn’t create 5 integer objects but creates only one integer object and all the indices of the array arr point to the same int object as shown. If we assign the 0th index to a another integer say 1, then a new integer object is created with the value of 1 and then the 0th index now points to this new int object as shown below
Strings in python are immutable, meaning no function can change the contents of an existing string, only provide a new string. Here's why.
See here for a discussion on string types that can be changed. In practice though, it's better to adjust to the immutability of strings.
List Manipulation in Python, List. A list is a collection which is ordered and changeable. In Python lists are written with This example returns the items from the beginning to "orange":. In Python programming, a list is created by placing all the items (elements) inside a square bracket [ ], separated by commas. It can have any number of items and they may be of different types (integer, float, string etc.). Also, a list can even have another list as an item. This is called nested list.
Python List, In this method, we just take the first element of list of tuple and the element at corresponding index and zip them together using the zip function. filter_none. edit Python : How to remove element from a list by value or Index | remove() vs pop() vs del; Python : Count elements in a list that satisfy certain conditions; Python : Check if a list contains all the elements of another list; Python : 6 Different ways to create Dictionaries; Find the index of value in Numpy Array using numpy.where()
Python list: how can I replace certain items of the list?, As a mutable, or changeable, ordered sequence of elements, lists are very flexible data structures in Python. List methods enable us to work with lists in a sophisticated manner. We can combine methods with other ways to modify lists in order to have a full range of tools to use lists effectively in our programs.
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- It does not work because strings are immutable.
re.subactually returns a string but you are not assigning it. Read the documentation: docs.python.org/library/re.html#re.sub
- List comprehension to the rescue!
- Is there any advantage of list comprehensions to standard for loops, or just readability or length of the code?
- @TooAngel: It is more awesome ;)
- @TooAngel: In addition to readability and reduced code length, list comprehensions also execute faster.
- @Daniel Stutzbach: Not always. It depends on what you are doing in the list comprehension (e.g. if you are using nested list comprehension). Sometimes you can gain a lot more performance using ordinary
- By default, I would go for this solution because you re-nest the new strings into the same list object. List comprehension approach detaches the result from original list pointer. When the list is property of a complex object, and you want that complex object to have the updated list, you will have to assign the new list back to a property, which is very un-pythonic.