如何使用舍入结果规范化数组（python、numpy、scipy）

此外，这不是正确的代码。 如何使它正确，简短和美丽

def normalize_weights(weights, threshold=0.01):
    total = sum(weights)
    result = [x / total for x in weights]
    result = [int((1.0 / threshold) * x) * threshold for x in result]
    result[-1] = 1.0 - sum(result[:-1])
    print(result)
    result[-1] = int((1.0 / threshold) * result[-1]) * threshold
    print(result)

normalize_weights([1.0, 1.0, 1.0])
[0.33, 0.33, 0.33999999999999997]
[0.33, 0.33, 0.34]  # ok

normalize_weights([1.0, 3.0, 1.0])
[0.2, 0.6, 0.19999999999999996]
[0.2, 0.6, 0.19]  # wrong

提前谢谢

---

编辑：结果之和应等于1.0

a = numpy.array([1.0,3.0,1.0])
normalized_a = numpy.round(a/numpy.linalg.norm(a,1.0),2)
# [0.2,0.6,0.2]

也许吧

a = numpy.array([1.0,1.0,1.0])
normalized_a = numpy.round(a/numpy.linalg.norm(a,1.0),2)
# [0.33,0.33,0.33]

如果你想保留2位小数，并强制设置为

1.0

，只需修改即可

normalized_a[0] += 1.0 - numpy.sum(normalized_a) # we could just as easily fix the -1 index ...

也许吧

a = numpy.array([1.0,1.0,1.0])
normalized_a = numpy.round(a/numpy.linalg.norm(a,1.0),2)
# [0.33,0.33,0.33]

如果你想保留2位小数，并强制设置为

1.0

，只需修改即可

normalized_a[0] += 1.0 - numpy.sum(normalized_a) # we could just as easily fix the -1 index ...

---

函数第二次调用的期望输出是什么？[0.2,0.6,0.2]结果之和应该等于1.0。函数第二次调用的期望输出是什么？[0.2,0.6,0.2]结果之和应该等于1.0。结果之和应该等于1.0。看起来不错。谢谢。结果之和应该等于1.0。看起来很棒。非常感谢。