在每个numpy数组元素上调用函数并在Python中覆盖其值

Question

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

Answer 1

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.

Answer 2

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

Answer 3

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 .