如何通过舍入结果来规范化数组(python、numpy、scipy)
How to normalize an array with rounding the result (python, numpy, scipy)
而且不是正确的代码。
如何做到正确、简短、美观?
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
提前致谢
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]
如果你想有 2 位小数并强制 1.0
只需修复它
normalized_a[0] += 1.0 - numpy.sum(normalized_a) # we could just as easily fix the -1 index ...
而且不是正确的代码。 如何做到正确、简短、美观?
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
提前致谢
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]
如果你想有 2 位小数并强制 1.0
只需修复它
normalized_a[0] += 1.0 - numpy.sum(normalized_a) # we could just as easily fix the -1 index ...