我有一个十进制数字列表,如下所示:
[-23.5, -12.7, -20.6, -11.3, -9.2, -4.5, 2, 8, 11, 15, 17, 21]
我需要将此列表规范化以适应范围[-5,5]
。
如何在Python中实现?
old_min = min(input)
old_range = max(input) - old_min
这里有一个棘手的部分。你可以用新范围乘以并用旧范围除以,但几乎可以保证顶桶中只会有一个值。你需要扩大输出范围,使得顶桶与其他所有桶的大小相同。
new_min = -5
new_range = 5 + 0.9999999999 - new_min
output = [floor((n - old_min) / old_range * new_range + new_min) for n in input]
0.9999999999
?如果有人有一个范围[0,1],那该数字会是多少呢? - Nairum>>> L = [-23.5, -12.7, -20.6, -11.3, -9.2, -4.5, 2, 8, 11, 15, 17, 21]
>>> normal = map(lambda x, r=float(L[-1] - L[0]): ((x - L[0]) / r)*10 - 5, L)
>>> normal
[-5.0, -2.5730337078651684, -4.348314606741574, -2.2584269662921352, -1.7865168539325844, -0.7303370786516856, 0.7303370786516847, 2.0786516853932575, 2.752808988764045, 3.6516853932584272, 4.101123595505618, 5.0]
original_vals = [-23.5, -12.7, -20.6, -11.3, -9.2, -4.5, 2, 8, 11, 15, 17, 21 ]
# get max absolute value
original_max = max([abs(val) for val in original_vals])
# normalize to desired range size
new_range_val = 5
normalized_vals = [float(val)/original_max * new_range_val for val in original_vals]
max
函数中不需要内部列表:max(abs(val) for val in original_vals)
也可以工作。 - Burhan Khalid# Rough code
# Get the range of the list
r = float(l[-1] - l[0])
# Normalize
normal = map(lambda x: (x - l[0]) / r, l)
基本上,您想将列表的基数调整为不同的范围。这将使您的原始列表归一化为[0,1]
保持简单:
>>> foo = [-23.5, -12.7, -20.6, -11.3, -9.2, -4.5, 2, 8, 11, 15, 17, 21]
>>> [i for i in foo if int(i) in range(-5, 5)]
[-4.5, 2]
另外,如果您希望结果仅为整数:
>>> [int(i) for i in foo if int(i) in range(-5, 5)]
[-4, 2]