我运行python神经网络预测,期待一个热门结果并获得如下数字:
[[0,1,0],
[0,0,1],
[0,1,0],
[1,0,0]
[0,1,0]]
如何将数组转换为一个热结果,即
let
答案 0 :(得分:2)
根据一个热门结果,我假设您希望每个子列表的最大值为1
,并且休息为0
(基于当前结果中的模式)。您可以使用 list comprehension 执行此操作:
>>> [[int(item == max(sublist)) else 0 for item in sublist] for sublist in my_list]
# ^ converts bool value returned by `==` into `int`. True -> 1, False -> 0
[[0, 1, 0], [0, 0, 1], [0, 1, 0], [1, 0, 0], [0, 1, 0]]
其中my_list
是您的初始列表。
但是在上面的方法中,每次迭代子列表时都会计算max()
。更好的方法就是:
def get_hot_value(my_list):
max_val = max(my_list)
return [int(item == max_val) for item in my_list]
hot_list = [get_hot_value(sublist) for sublist in my_list]
修改:如果您应该在列表中只有一个1
(如果超过1个元素的最大值),您可以修改get_hot_value
功能如:
def get_hot_value(my_list):
max_val, hot_list, is_max_found = max(my_list), [], False
for item in my_list:
if item == max_val and not is_max_found:
hot_list.append(1)
else:
hot_list.append(0)
is_max_found = True
return hot_list
答案 1 :(得分:1)
我会建议:
n = [[0.33058667182922363, 0.3436272442340851, 0.3257860243320465],
[0.32983461022377014, 0.3487854599952698, 0.4213798701763153],
[0.3311253488063812, 0.3473075330257416, 0.3215670585632324],
[0.38368630170822144, 0.35151687264442444, 0.3247968554496765],
[0.3332786560058594, 0.343686580657959, 0.32303473353385925]]
hot_results = []
for row in n:
hot_index = row.index(max(row))
hot_result = [0] * len(row)
hot_result[hot_index] = 1
hot_results.append(hot_result)
print(hot_results)
答案 2 :(得分:1)
其他解决方案都很好,并解决了问题。或者,如果你有numpy,
import numpy as np
n = [[0.33058667182922363, 0.3436272442340851, 0.3257860243320465],
[0.32983461022377014, 0.3487854599952698, 0.4213798701763153],
[0.3311253488063812, 0.3473075330257416, 0.3215670585632324],
[0.38368630170822144, 0.35151687264442444, 0.3247968554496765],
[0.3332786560058594, 0.343686580657959, 0.32303473353385925]]
max_indices = np.argmax(n,axis=1)
final_values = [n[i] for i in max_indices]
argmax能够找到该行中最大值的索引,然后你只需要对其进行一次列表理解。应该是相当快的我猜?