How can I remove an element from a dictionary based on the index? For example, if I have
d = {'s':0, 'e':1, 't':6}
I want to remove and return the first entry (leftmost), 's' and 0, such that the dictionary has
d = {'e':1, 't':6}
I have tried d.popitem() that removes 't':6. d.pop(key) removes a key from the dictionary, but I have no way of finding the key I want to remove, I only have the index.
Assuming you're doing this in Python 3.7 or later, where dict keys are ordered, you can create an iterator from the dict keys and use the next function to obtain the first key, so that you can use the del statement to remove that key:
d = {'s':0, 'e':1, 't':6}
del d[next(iter(d))]
print(d)
This outputs:
{'e': 1, 't': 6}
If you want to remove a key of a different index, you can use itertools.islice to obtain the key at a given index. For example, to delete the second key (of index 1) from d:
from itertools import islice
d = {'s':0, 'e':1, 't':6}
del d[next(islice(d, 1, None))]
print(d)
This outputs:
{'s': 0, 't': 6}
del d[list(d)[index]]
list constructor you're forcing iterations over all keys of the dict, whereas with islice it can short-circuit once the desired index is reached.del d[next(islice(d, 50, None))]Quick 1 line code:
myDict = {"a":1, "b":2, "c":3}
indexToBeRemoved = 1
myDict.pop(list(myDict.keys())[indexToBeRemoved])
Had the same problem. idk if its right but it works on my code
grades = {
"lance": [1, 2, 3],
"nate": [1, 2, 3]
}
index = 0
keys = list(grades)
key = keys[index]
grades.pop(key)
print(grades)
You can try converting the dict temporarily into a list and then popping like this:
for ele in list(dict1)[0]:
dict1.pop(ele)
dict = {'s':0, 'e':1, 't':6}
# del dict['key']
#Option 1
del dict['s']
print (dict)
#Option 2
new_dict = dict.pop('s')
print(dict)
you can do this by iterating your dictionary.
1) if you have only one index to delete
i = 0
for key in d.keys():
if i == 1: #assuming you want to remove the second element from the dictionary
key_to_delete = key
i = i + 1
if key_to_delete in d:
del d[key_to_delete]
print(d)
2) if you have multiple indexes to delete:
i = 0
index_to_delete = [1,2] #assuming you want to remove the second and the third elements from the dictionary
keys_to_delete = []
for key in d.keys():
if i in index_to_delete:
keys_to_delete.append(key)
i = i + 1
for key in keys_to_delete:
if key in d:
del d[key]
print(d)
if key_to_delete in d? That doesn't prevent a race condition, if that's what you had in mind (though it at least makes it smaller). Instead, put del d[key_to_delete] in a try block and except KeyError if you want to properly handle the key not existing. In this case that would be pass, but in production code, you should probably at least log it.i = i + 1 should be i += 1. 2) To loop over indices and elements, use enumerate: for i, key in enumerate(d.keys()):.if i == 1), you should break the loop to avoid looping more than necessary. Otherwise, you might as well convert to list and select element 1: list(d.keys())[1]. 4) To get multiple elements from a list, use operator.itemgetter(*items). 5) If you have a data structure whose only purpose is membership checks, i.e. index_to_delete, use a set to get O(1) (average) lookup time.
list(my_dict)[index]