You can use a list comprehension with enumerate:
indices = [i for i, x in enumerate(my_list) if x == "whatever"]
The iterator enumerate(my_list) yields pairs (index, item) for each item in the list. Using i, x as loop variable target unpacks these pairs into the index i and the list item x. We filter down to all x that match our criterion, and select the indices i of these elements.
You can use a list comprehension with enumerate:
indices = [i for i, x in enumerate(my_list) if x == "whatever"]
The iterator enumerate(my_list) yields pairs (index, item) for each item in the list. Using i, x as loop variable target unpacks these pairs into the index i and the list item x. We filter down to all x that match our criterion, and select the indices i of these elements.
While not a solution for lists directly, numpy really shines for this sort of thing:
import numpy as np
values = np.array([1,2,3,1,2,4,5,6,3,2,1])
searchval = 3
ii = np.where(values == searchval)[0]
returns:
ii ==>array([2, 8])
This can be significantly faster for lists (arrays) with a large number of elements vs some of the other solutions.
python - Find a value in a list - Stack Overflow
Help with .index()? finding multiple instances of item
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As for your first question: "if item is in my_list:" is perfectly fine and should work if item equals one of the elements inside my_list. The item must exactly match an item in the list. For instance, "abc" and "ABC" do not match. Floating point values in particular may suffer from inaccuracy. For instance, 1 - 1/3 != 2/3.
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]
If you want to find one element or None use default in next, it won't raise StopIteration if the item was not found in the list:
first_or_default = next((x for x in lst if ...), None)
Taken from https://www.programiz.com/python-programming/online-compiler/?ref=409055e9 :
vowels = ['a', 'e', 'i', 'o', 'i', 'u']
# index of the first 'i' is returned
index = vowels.index('i')
print('The index of i:', index)
Output: The index of i: 2
Say that the list was much bigger, and you don't know the contents, but you know 'i' is in it more than once. What would be the best way to find all instances of 'i'?
Thanks so much! <3