What is the most ‘pythonic’ way of finding an item in a list? Is it:
if item in my_list:
print("Desired item is in list")
13 s
As for your first question: that code is perfectly fine and should work if item equals one of the elements inside myList. Maybe you try to find a string that does not exactly match one of the items or maybe you are using a float value which suffers from inaccuracy.
As for your second question: There’s actually several possible ways if “finding” things in lists.
Checking if something is inside
This is the use case you describe: Checking whether something is inside a list or not. As you know, you can use the in operator for that:
3 in [1, 2, 3] # => True
Filtering a collection
That is, finding all elements in a sequence that meet a certain condition. You can use list comprehension or generator expressions for that:
matches = [x for x in lst if fulfills_some_condition(x)]
matches = (x for x in lst if x > 6)
The latter will return a generator which you can imagine as a sort of lazy list that will only be built as soon as you iterate through it. By the way, the first one is exactly equivalent to
matches = filter(fulfills_some_condition, lst)
in Python 2. Here you can see higher-order functions at work. In Python 3, filter doesn’t return a list, but a generator-like object.
Finding the first occurrence
If you only want the first thing that matches a condition (but you don’t know what it is yet), it’s fine to use a for loop (possibly using the else clause as well, which is not really well-known). You can also use
next(x for x in lst if ...)
which will return the first match or raise a StopIteration if none is found. Alternatively, you can use
next((x for x in lst if ...), [default value])
Finding the location of an item
For lists, there’s also the index method that can sometimes be useful if you want to know where a certain element is in the list:
[1,2,3].index(2) # => 1
[1,2,3].index(4) # => ValueError
However, note that if you have duplicates, .index always returns the lowest index:……
[1,2,3,2].index(2) # => 1
If there are duplicates and you want all the indexes then you can use enumerate() instead:
[i for i,x in enumerate([1,2,3,2]) if x==2] # => [1, 3]