If a value already exists in a list, I want to remove it (which it may not).
a = [1, 2, 3, 4] b = a.index(6) del a[b] print(a)
The following error is shown in the above scenario (when it does not exist):
Traceback (most recent call last): File "D:\zjm_code\a.py", line 6, in <module> b = a.index(6) ValueError: list.index(x): x not in list
So here’s what I have to do:
a = [1, 2, 3, 4] try: b = a.index(6) del a[b] except: pass print(a)
Isn’t there a more straightforward approach?
Asked by zjm1126
Simply use list to eliminate an element’s initial appearance in a list. remove:
>>> a = ['a', 'b', 'c', 'd'] >>> a.remove('b') >>> print(a) ['a', 'c', 'd']
Keep in mind that it does not eliminate all instances of your element. To do so, use a list comprehension exercise.
>>> a = [10, 20, 30, 40, 20, 30, 40, 20, 70, 20] >>> a = [x for x in a if x != 20] >>> print(a) [10, 30, 40, 30, 40, 70]
Answered by Johannes Charra
If you ask Python to do something it can’t, it will usually throw an Exception, therefore you’ll have to perform one of two things:
if c in a: a.remove(c)
try: a.remove(c) except ValueError: pass
An exception isn’t always a terrible thing, as long as you anticipate it and handle it properly.
Answered by Dave Webb
You can do
a=[1,2,3,4] if 6 in a: a.remove(6)
However, because searching 6 items in the list twice is inconvenient, trying unless would be a better option.
try: a.remove(6) except: pass
Answered by YOU
a = [1,2,2,3,4,5]
To take out all occurrences, you could use the filter function in python. For instance, consider the following:
a = list(filter(lambda x: x!= 2, a))
As a result, all elements of a!= 2 would be preserved.
To simply remove one of the elements, use
Answered by mathwizurd
Here’s how to accomplish it in real time (without using a list):
def remove_all(seq, value): pos = 0 for item in seq: if item != value: seq[pos] = item pos += 1 del seq[pos:]
Answered by jfs
Post is based on https://stackoverflow.com/questions/2793324/is-there-a-simple-way-to-delete-a-list-element-by-value