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Python: Remove an item from a set if it is present in the set

Python sets: Exercise-5 with Solution

Write a Python program to remove an item from a set if it is present in the set.

Sample Solution:

Python Code:

#Create a new set
num_set = set([0, 1, 2, 3, 4, 5])
print("Original set elements:")
print(num_set)
print("\nRemove 0 from the said set:")
num_set.discard(4)
print(num_set)
print("\nRemove 5 from the said set:")
num_set.discard(5)
print(num_set)
print("\nRemove 2 from the said set:")
num_set.discard(5)
print(num_set)
print("\nRemove 7 from the said set:")
num_set.discard(15)
print(num_set)

Sample Output:

Original set elements:
{0, 1, 2, 3, 4, 5}

Remove 0 from the said set:
{0, 1, 2, 3, 5}

Remove 5 from the said set:
{0, 1, 2, 3}

Remove 2 from the said set:
{0, 1, 2, 3}

Remove 7 from the said set:
{0, 1, 2, 3}

Pictorial Presentation:

NumPy Sets: Remove an item from a set if it is present in the set.

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Python: Tips of the Day

Find the index of an item in a list?

Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python?

>>> ["foo", "bar", "baz"].index("bar")
1

Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can't remember the last time I used it in anger. It's been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:

list.index(x[, start[, end]])

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If your list is long, and you don't know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:

>>> import timeit
>>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
9.356267921015387
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
0.0004404920036904514

Only returns the index of the first match to its argument

A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2

Most places where I once would have used index, I now use a list comprehension or generator expression because they're more generalizable. So if you're considering reaching for index, take a look at these excellent Python features.

Throws if element not present in list

A call to index results in a ValueError if the item's not present.

>>> [1, 1].index(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 2 is not in list

If the item might not be present in the list, you should either

  • Check for it first with item in my_list (clean, readable approach), or
  • Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)

Ref: https://bit.ly/2ALwXwe