PYTHON

python get index of first element of list that matches condition

# Find first index (or value) that meets condition
my_list = [2, 4, 6, 9, 10]

# Option 1. Use an iterator (does not scan all list)
y = (i for i,x in enumerate(my_list) if is_odd(x))
idx1 = next(y)  # <== index of first element

y = (x for i,x in enumerate(my_list) if is_odd(x))
v1 = next(y)  # <== value of first element


# Option 2. Use a list comprehension (scans all list)
idx1 = [i for i,x in enumerate(my_list) if x % 2 != 0][0]
v1 = [x for i,x in enumerate(my_list) if x % 2 != 0][0]

python find index of first matching element in a list

# Basic syntax:
list.index(element, start, end) 
# Where:
#	- Element is the item you're looking for in the list
# 	- Start is optional, and is the list index you want to start at
#	- End is optional, and is the list index you want to stop searching at

# Note, Python is 0-indexed

# Example usage:
my_list = [1, 2, 3, 4, 5, 6, 7, 8, 42, 9, 10]
my_list.index(42)
--> 8