在Python中,我需要生成一个将字母映射到该字母的预定义“ 单热 ”表示的dict 。 举例来说, dict应该是这样的:
{ 'A': '1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0', 'B': '0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0', # ... }每个字母表中有一位(表示为字符)。 因此,每个字符串将包含25个零和一个1. 1的位置由字母表中相应字母的位置决定。
我想出了一些产生这种情况的代码:
# Character set is explicitly specified for fine grained control _letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ" n = len(_letters) one_hot = [' '.join(['0']*a + ['1'] + ['0']*b) for a, b in zip(range(n), range(n-1, -1, -1))] outputs = dict(zip(_letters, one_hot))是否有更高效/更清洁/更pythonic的方式来做同样的事情?
In Python, I need to generate a dict that maps a letter to a pre-defined "one-hot" representation of that letter. By way of illustration, the dict should look like this:
{ 'A': '1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0', 'B': '0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0', # ... }There is one bit (represented as a character) per letter of the alphabet. Hence each string will contain 25 zeros and one 1. The position of the 1 is determined by the position of the corresponding letter in the alphabet.
I came up with some code that generates this:
# Character set is explicitly specified for fine grained control _letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ" n = len(_letters) one_hot = [' '.join(['0']*a + ['1'] + ['0']*b) for a, b in zip(range(n), range(n-1, -1, -1))] outputs = dict(zip(_letters, one_hot))Is there a more efficient/cleaner/more pythonic way to do the same thing?
最满意答案
我觉得这更可读:
from string import ascii_uppercase one_hot = {} for i, l in enumerate(ascii_uppercase): bits = ['0']*26; bits[i] = '1' one_hot[l] = ' '.join(bits)如果您需要更一般的字母表,只需枚举字符串,并用['0']*len(alphabet)替换['0']*26 。
I find this to be more readable:
from string import ascii_uppercase one_hot = {} for i, l in enumerate(ascii_uppercase): bits = ['0']*26; bits[i] = '1' one_hot[l] = ' '.join(bits)If you need a more general alphabet, just enumerate over a string of the characters, and replace ['0']*26 with ['0']*len(alphabet).
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