将2d整数数组转换为字符串python

Question

convert a 2d array of integers into String. For example myArray = [[1,2,3],[3,4,5]] into myArrayL = [[a,b,c],[c,d,e]] and retain the shape of the 2d array

where 1 = "a" , 2 = "b", 3 = "c"

Answer 1

There are two parts to this.

First, how do you convert a number like 1 to a letter like 'a' , according to your rule? Second, how do you apply a function to all elements of a list?

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For the first, one way to write it is with the chr function. This function takes an number and gives you the single-character string for the character with that code point. In particular, chr(65) is 'a' , chr(66) is 'b' , etc. So, we could just do chr(n + 64) .

Or we could use the ord function, which is the inverse of chr , so instead of hardcoding 64 and having to remember that's 1 less than 'a' , we can write 1 less than 'a' directly:

def letter(n):
    return chr(n + ord('a') - 1)

Of course this isn't the only way to do it. You could also, eg, use string.ascii_lowercase[n-1] .

(Note that either of these solutions not only works in Python 3, where the "code points" are always Unicode, but also in Python 2, where the "code points" are values in some unspecific 8-bit encoding, as long as the intended encoding has all of the lowercase letters in contiguous order, which is true for almost anything you're likely to ever encounter unless you've got some old EBCDIC files lying around.)

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For the second, you can use a list comprehension . Your examples are flat (1D) lists, so we'd use a flat list comprehension:

numbers = [2, 3, 4]
letters = [letter(n) for n in numbers]

If you have 2D lists of lists, just use a nested list comprehension :

numbers = [[2, 3], [4, 5]]
letters = [[letter(n) for n in row] for row in numbers]

Answer 2

You can corresponding Alphabets by indexing string.ascii_lowercase which returns all lower case alphabets.

import string 
myArray = [[1,2,3],[3,4,5]]
result_array = [[string.ascii_lowercase[element-1] for element in row] for row in myArray]

Result array:

[['a', 'b', 'c'], ['c', 'd', 'e']]