In Python, how can I remove an object from array of objects? Like this:
x = object()
y = object()
array = [x,y]
# Remove x
I've tried array.remove() but it only works with a value, not a specific location in the array. I need to be able to delete the object by addressing its position(remove array[0])
You could try this to dynamically remove an object from an array without looping through it? Where e and t are just random objects.
>>> e = {'b':1, 'w': 2}
>>> t = {'b':1, 'w': 3}
>>> p = [e,t]
>>> p
[{'b': 1, 'w': 2}, {'b': 1, 'w': 3}]
>>>
>>> p.pop(p.index({'b':1, 'w': 3}))
{'b': 1, 'w': 3}
>>> p
[{'b': 1, 'w': 2}]
>>>
If you know the array location you can can pass it into itself. If you are removing multiple items I suggest you remove them in reverse order.
#Setup array
array = [55,126,555,2,36]
#Remove 55 which is in position 0
array.remove(array[0])
If you want to remove multiple object from a list. There are various ways to delete an object from a list
Try this code. a is list with all object, b is list object you want to remove.
example :
a = [1,2,3,4,5,6]
b = [2,3]
for i in b:
if i in a:
a.remove(i)
print(a)
the output is [1,4,5,6]
I hope, it will work for you
if you wanna remove the last one just do your_list.pop(-1)
if you wanna remove the first one your_list.pop(0) or any index you wish to remove
In python there are no arrays, lists are used instead. There are various ways to delete an object from a list:
my_list = [1,2,4,6,7]
del my_list[1] # Removes index 1 from the list
print my_list # [1,4,6,7]
my_list.remove(4) # Removes the integer 4 from the list, not the index 4
print my_list # [1,6,7]
my_list.pop(2) # Removes index 2 from the list
In your case the appropriate method to use is pop, because it takes the index to be removed:
x = object()
y = object()
array = [x, y]
array.pop(0)
# Using the del statement
del array[0]
del array[0]
where 0 is the index of the object in the list (there is no array in python)
You can remove a string from an array like this:
array = ["Bob", "Same"]
array.remove("Bob")
Source: Stackoverflow.com