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如何将数字四舍五入到指定的上限或下限?

upperbound lower-bound rounding integer python

P.hunter · 技术社区 · 7 年前

我正在处理一个数据集,在这个数据集上我有一些值需要四舍五入到上下限。

9 降到 3. 我们有这样的数字-

[ 7.453511737983394, 
  8.10917072790058, 
  6.2377799380575, 
  5.225853201122676, 
  4.067932296134156 ]

我们希望列表四舍五入到3或9,就像这样-

[ 9, 
  9, 
  9, 
  3, 
  3 ]

我知道我们可以用一种很好的老方法来做,比如在数组中迭代,找到差异,然后得到最接近的一个。

我的方法代码:

for i in the_list[:]:
    three = abs(3-the_list[i])  
    nine = abs(9-the_list[i])

    if three < nine:
        the_list[i] = three
    else:
        the_list[i] = nine

又快又脏 内置于python中的方式,如:

hey_bound = round_the_num(number, bound_1, bound_2)

我知道我们可以 my-approach-code

8 回复 | 直到 7 年前

SpghttCd 7 年前

编辑:
the_list 和两个界限(因此这里没有昂贵的乘法),然后仅有条件地加上一个或另一个,取决于哪个更小:

import numpy as np

the_list = np.array([ 7.453511737983394,
8.10917072790058, 
6.2377799380575, 
5.225853201122676, 
4.067932296134156 ])

dhi = 9 - the_list
dlo = 3 - the_list
idx = dhi + dlo < 0
the_rounded = the_list + np.where(idx, dhi, dlo)
# array([9., 9., 9., 3., 3.])

我将对无偏移的规范化列表应用round函数,然后缩小并添加偏移:

import numpy as np

the_list = np.array([ 7.453511737983394,
8.10917072790058, 
6.2377799380575, 
5.225853201122676, 
4.067932296134156 ])

hi = 9
lo = 3
dlt = hi - lo

the_rounded = np.round((the_list - lo)/dlt) * dlt + lo

# [9. 9. 9. 3. 3.]

SpghttCd 7 年前

定时比较 现有答案的数量

我的解释是:
从性能的角度来看,你应该选择Abhishek Patel或Carles Mitjans来获得较小的列表。
对于包含几十个或更多值的列表,numpy数组然后有条件地添加与较小绝对值的差异似乎是最快的解决方案。

import timeit
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

plt.style.use('ggplot')

rep = 5

timings = dict()

for n in range(7):
    print(f'N = 10^{n}')

    N = 10**n
    setup = f'''import numpy as np\nthe_list = np.random.random({N})*6+3\nhi = 9\nlo = 3\ndlt = hi - lo\nmid = (hi + lo) / 2\ndef return_the_num(l, lst, h):\n    return [l if abs(l-x) < abs(h-x) else h for x in lst]'''

    fct = 'np.round((the_list - lo)/dlt) * dlt + lo'
    t = timeit.Timer(fct, setup=setup)
    timings['SpghttCd_np'] = timings.get('SpghttCd_np', []) + [np.min(t.repeat(repeat=rep, number=1))]

    fct = 'return_the_num(3, the_list, 9)'
    t = timeit.Timer(fct, setup=setup)
    timings['Austin'] = timings.get('Austin', []) + [np.min(t.repeat(repeat=rep, number=1))]

    fct = '[(lo, hi)[mid < v] for v in the_list]'
    t = timeit.Timer(fct, setup=setup)
    timings['SpghttCd_lc'] = timings.get('SpghttCd_lc', []) + [np.min(t.repeat(repeat=rep, number=1))]

    setup += '\nround_the_num = lambda list, upper, lower: [upper if x > (upper + lower) / 2 else lower for x in list]'
    fct = 'round_the_num(the_list, 9, 3)'
    t = timeit.Timer(fct, setup=setup)
    timings['Carles Mitjans'] = timings.get('Carles Mitjans', []) + [np.min(t.repeat(repeat=rep, number=1))]

    setup += '\nupper_lower_bound_list=[3,9]'
    fct = '[min(upper_lower_bound_list, key=lambda x:abs(x-myNumber)) for myNumber in the_list]'
    t = timeit.Timer(fct, setup=setup)
    timings['mad_'] = timings.get('mad_', []) + [np.min(t.repeat(repeat=rep, number=1))]

    setup += '\ndef return_bound(x, l, h):\n    low = abs(x - l)\n    high = abs(x - h)\n    if low < high:\n        return l\n    else:\n        return h'
    fct = '[return_bound(x, 3, 9) for x in the_list]'
    t = timeit.Timer(fct, setup=setup)
    timings["Scratch'N'Purr"] = timings.get("Scratch'N'Purr", []) + [np.min(t.repeat(repeat=rep, number=1))]

    setup += '\ndef round_the_list(list, bound_1, bound_2):\n\tmid = (bound_1+bound_2)/2\n\tfor i in range(len(list)):\n\t\tif list[i] > mid:\n\t\t\tlist[i] = bound_2\n\t\telse:\n\t\t\tlist[i] = bound_1'
    fct = 'round_the_list(the_list, 3, 9)'
    t = timeit.Timer(fct, setup=setup)
    timings["Abhishek Patel"] = timings.get("Abhishek Patel", []) + [np.min(t.repeat(repeat=rep, number=1))]

    fct = 'dhi = 9 - the_list\ndlo = 3 - the_list\nidx = dhi + dlo < 0\nthe_list + np.where(idx, dhi, dlo)'
    t = timeit.Timer(fct, setup=setup)
    timings["SpghttCd_where"] = timings.get("SpghttCd_where", []) + [np.min(t.repeat(repeat=rep, number=1))]

print('done')

df = pd.DataFrame(timings, 10**np.arange(n+1))
ax = df.plot(logx=True, logy=True)
ax.set_xlabel('length of the list')
ax.set_ylabel('seconds to run')
ax.get_lines()[-1].set_c('g')
plt.legend()
print(df)

Abhishek Patel 7 年前

您可以通过找到中点并检查列表中每个数字位于中点的哪一侧来进行概括

def round_the_list(list, bound_1, bound_2):
  mid = (bound_1+bound_2)/2
  for i in range(len(list)):
        if list[i] > mid:         # or >= depending on your rounding decision
            list[i] = bound_2
        else:
            list[i] = bound_1

Scratch'N'Purr 7 年前

也许您可以编写一个函数并在列表理解中使用它。

def return_bound(x, l, h):
    low = abs(x - l)
    high = abs(x - h)
    if low < high:
        return l
    else:
        return h

测试:

>>> mylist = [7.453511737983394, 8.10917072790058, 6.2377799380575, 5.225853201122676, 4.067932296134156]
>>> [return_bound(x, 3, 9) for x in mylist]
[9, 9, 9, 3, 3]

mad_ 7 年前

使用内置语言的一行列表理解 min 函数通过修改键参数来查找绝对差异

upper_lower_bound_list=[3,9]
myNumberlist=[ 7.453511737983394, 
8.10917072790058, 
6.2377799380575, 
5.225853201122676, 
4.067932296134156 ]

列表理解

[min(upper_lower_bound_list, key=lambda x:abs(x-myNumber)) for myNumber in myNumberlist]

输出

[9, 9, 9, 3, 3]

Carles Mitjans 7 年前

使用列表理解和lambda函数的另一个选项:

round_the_num = lambda list, upper, lower: [upper if x > (upper + lower) / 2 else lower for x in list]

round_the_num(l, 9, 3)

Austin 7 年前

您可以编写执行列表理解的自定义函数,如:

lst = [ 7.453511737983394, 
  8.10917072790058, 
  6.2377799380575, 
  5.225853201122676, 
  4.067932296134156 ]

def return_the_num(l, lst, h): 
    return [l if abs(l-x) < abs(h-x) else h for x in lst]

print(return_the_num(3, lst, 9))
# [9, 9, 9, 3, 3]

SpghttCd 7 年前

我真的很喜欢@AbhishekPatel关于与中点进行比较的想法。但我将其放入LC,使用结果作为边界元组的索引:

the_list = [ 7.453511737983394,
8.10917072790058, 
6.2377799380575, 
5.225853201122676, 
4.067932296134156 ]

hi = 9
lo = 3
mid = (hi + lo) / 2

[(lo, hi)[mid < v] for v in the_list]
# [9, 9, 9, 3, 3]

然而,这里可以处理大于 hi 或低于 lo .