I have a list:
letters = ('1', '2', '3', '5', '8', '13', '21')
Having number 8, I can get it's index in letters as:
letters.index('8')
No, I have an array which contains only numbers that are in letters.
numbers = [3, 8, 13, 21, 5]
How do I call index (or similar) function on each element?
what I want to get is [2, 4, 5, 6, 3]
Is there such function as index that can grab array instead on single element? If that is required letters can be changed to array
You can try
letters = ('1', '2', '3', '5', '8', '13', '21')
numbers = [3, 8, 13, 21, 5]
result = [letters.index(str(n)) for n in numbers]
But this works only if all requests are present in letters
You can try this to test if the request in letters
[letters.index(str(n)) if str(n) in letters else -1 for n in numbers]
In case request in not letters, you will have a -1. Beware that -1 is here to indicate that request is not present. letters[-1] returns the last element of the tuple.
I like to use map functions - this is a great problem to apply a new technique.
def get_index(letters, element):
return letters.index(str(element))
indices_of_numbers = list(map(get_index, numbers))
we could wrap this in a function so we can remove the letters argument from get_index.
def get_indices_of_numbers(letters, numbers):
def get_index(element):
return letters.index(str(element))
indices_of_numbers = list(map(get_index, numbers))
return indices_of_numbers
You could convert letters to int array and then use np.searchsorted -
np.asarray(letters,dtype=int).searchsorted(numbers)
Sample run -
In [42]: letters = ('1', '2', '3', '5', '8', '13', '21')
In [43]: numbers = [3, 8, 13, 21, 5]
In [44]: np.asarray(letters,dtype=int).searchsorted(numbers)
Out[44]: array([2, 4, 5, 6, 3])
If letters is such that the int array version is not sorted, we need to feed in the additional argument sorter with searchsorted.
lettersis a tuple, not a list.numbersis a list, not an array.