Question

编辑：由于我的代码中的错误，我使用我最早但工作正常的代码进行了更新

我从数据库中获得了速度记录列表，我想在该列表中找到最大速度。听起来很容易，但我对任何最大速度都有一些要求：

如果最大速度超过一定水平，则必须将超过一定数量的记录识别为最大速度。这种逻辑的原因是我想要在正常条件下的最大速度，而不仅仅是错误或一次出现。由于同样的原因，我还有一个限制，即速度必须超过一定的限制。

以下是速度阵列的示例：

v = [8.0, 1.3, 0.7, 0.8, 0.9, 1.1, 14.9, 14.0, 14.1, 14.2, 14.3, 13.8, 13.9, 13.7, 13.6, 13.5, 13.4, 15.7, 15.8, 15.0, 15.3, 15.4, 15.5, 15.6, 15.2, 12.8, 12.7, 12.6, 8.7, 8.8, 8.6, 9.0, 8.5, 8.4, 8.3, 0.1, 0.0, 16.4, 16.5, 16.7, 16.8, 17.0, 17.1, 17.8, 17.7, 17.6, 17.4, 17.5, 17.3, 17.9, 18.2, 18.3, 18.1, 18.0, 18.4, 18.5, 18.6, 19.0, 19.1, 18.9, 19.2, 19.3, 19.9, 20.1, 19.8, 20.0, 19.7, 19.6, 19.5, 20.2, 20.3, 18.7, 18.8, 17.2, 16.9, 11.5, 11.2, 11.3, 11.4, 7.1, 12.9, 14.4, 13.1, 13.2, 12.5, 12.1, 12.2, 13.0, 0.2, 3.6, 7.4, 4.6, 4.5, 4.3, 4.0, 9.4, 9.6, 9.7, 5.8, 5.7, 7.3, 2.1, 0.4, 0.3, 16.1, 11.9, 12.0, 11.7, 11.8, 10.0, 10.1, 9.8, 15.1, 14.7, 14.8, 10.2, 10.3, 1.2, 9.9, 1.9, 3.4, 14.6, 0.6, 5.1, 5.2, 7.5, 19.4, 10.7, 10.8, 10.9, 0.5, 16.3, 16.2, 16.0, 16.6, 12.4, 11.0, 1.7, 1.6, 2.4, 11.6, 3.9, 3.8, 14.5, 11.1]

这是我的代码，可以找到我定义的真正最大速度：

from collections import Counter
while max(speeds)>30:
    speeds.remove(max(speeds))
nwsp = []
for s in speeds:
    nwsp.append(np.floor(s))
count = Counter(nwsp)
while speeds and max(speeds)>14 and count[np.floor(max(speeds))]<10:
    speeds.remove(max(speeds))
while speeds and max(speeds)<5:
    speeds.remove(max(speeds))
if speeds:
    print max(speeds)
    return max(speeds)
else:
    return False

结果如图所示v：19.9

我制作nwsp的原因是，如果f.ex 19.6只被发现9次对我来说无关紧要 - 如果在同一个整数内有任何数字，f.ex 19.7也被找到3次，那么19.6将有效。

如何重写/优化此代码，以便选择过程更快？我已经删除了max（速度），而是对列表进行了排序，并使用velocity [-1]引用了最大的元素。

很抱歉没有为我的速度添加任何单位。

Answer 1

您的代码速度很慢，因为您一遍又一遍地调用max和remove，并且每次调用都会花费与列表长度成比例的时间。任何合理的解决方案都会快得多。

如果您知道False无法发生，那么这就足够了：

speeds = [8.0, 1.3, 0.7, 0.8, 0.9, 1.1, 14.9, 14.0, 14.1, 14.2, 14.3, 13.8, 13.9, 13.7, 13.6, 13.5, 13.4, 15.7, 15.8, 15.0, 15.3, 15.4, 15.5, 15.6, 15.2, 12.8, 12.7, 12.6, 8.7, 8.8, 8.6, 9.0, 8.5, 8.4, 8.3, 0.1, 0.0, 16.4, 16.5, 16.7, 16.8, 17.0, 17.1, 17.8, 17.7, 17.6, 17.4, 17.5, 17.3, 17.9, 18.2, 18.3, 18.1, 18.0, 18.4, 18.5, 18.6, 19.0, 19.1, 18.9, 19.2, 19.3, 19.9, 20.1, 19.8, 20.0, 19.7, 19.6, 19.5, 20.2, 20.3, 18.7, 18.8, 17.2, 16.9, 11.5, 11.2, 11.3, 11.4, 7.1, 12.9, 14.4, 13.1, 13.2, 12.5, 12.1, 12.2, 13.0, 0.2, 3.6, 7.4, 4.6, 4.5, 4.3, 4.0, 9.4, 9.6, 9.7, 5.8, 5.7, 7.3, 2.1, 0.4, 0.3, 16.1, 11.9, 12.0, 11.7, 11.8, 10.0, 10.1, 9.8, 15.1, 14.7, 14.8, 10.2, 10.3, 1.2, 9.9, 1.9, 3.4, 14.6, 0.6, 5.1, 5.2, 7.5, 19.4, 10.7, 10.8, 10.9, 0.5, 16.3, 16.2, 16.0, 16.6, 12.4, 11.0, 1.7, 1.6, 2.4, 11.6, 3.9, 3.8, 14.5, 11.1]

from collections import Counter
count = Counter(map(int, speeds))
print max(s for s in speeds
          if 5 <= s <= 30 and (s <= 14 or count[int(s)] >= 10))

如果False案件可能发生，这将是一种方式：

speeds = [8.0, 1.3, 0.7, 0.8, 0.9, 1.1, 14.9, 14.0, 14.1, 14.2, 14.3, 13.8, 13.9, 13.7, 13.6, 13.5, 13.4, 15.7, 15.8, 15.0, 15.3, 15.4, 15.5, 15.6, 15.2, 12.8, 12.7, 12.6, 8.7, 8.8, 8.6, 9.0, 8.5, 8.4, 8.3, 0.1, 0.0, 16.4, 16.5, 16.7, 16.8, 17.0, 17.1, 17.8, 17.7, 17.6, 17.4, 17.5, 17.3, 17.9, 18.2, 18.3, 18.1, 18.0, 18.4, 18.5, 18.6, 19.0, 19.1, 18.9, 19.2, 19.3, 19.9, 20.1, 19.8, 20.0, 19.7, 19.6, 19.5, 20.2, 20.3, 18.7, 18.8, 17.2, 16.9, 11.5, 11.2, 11.3, 11.4, 7.1, 12.9, 14.4, 13.1, 13.2, 12.5, 12.1, 12.2, 13.0, 0.2, 3.6, 7.4, 4.6, 4.5, 4.3, 4.0, 9.4, 9.6, 9.7, 5.8, 5.7, 7.3, 2.1, 0.4, 0.3, 16.1, 11.9, 12.0, 11.7, 11.8, 10.0, 10.1, 9.8, 15.1, 14.7, 14.8, 10.2, 10.3, 1.2, 9.9, 1.9, 3.4, 14.6, 0.6, 5.1, 5.2, 7.5, 19.4, 10.7, 10.8, 10.9, 0.5, 16.3, 16.2, 16.0, 16.6, 12.4, 11.0, 1.7, 1.6, 2.4, 11.6, 3.9, 3.8, 14.5, 11.1]

from collections import Counter
count = Counter(map(int, speeds))
valids = [s for s in speeds
         if 5 <= s <= 30 and (s <= 14 or count[int(s)] >= 10)]
print max(valids) if valids else False

或者排序并使用next，这可以将您的False作为默认值：

speeds = [8.0, 1.3, 0.7, 0.8, 0.9, 1.1, 14.9, 14.0, 14.1, 14.2, 14.3, 13.8, 13.9, 13.7, 13.6, 13.5, 13.4, 15.7, 15.8, 15.0, 15.3, 15.4, 15.5, 15.6, 15.2, 12.8, 12.7, 12.6, 8.7, 8.8, 8.6, 9.0, 8.5, 8.4, 8.3, 0.1, 0.0, 16.4, 16.5, 16.7, 16.8, 17.0, 17.1, 17.8, 17.7, 17.6, 17.4, 17.5, 17.3, 17.9, 18.2, 18.3, 18.1, 18.0, 18.4, 18.5, 18.6, 19.0, 19.1, 18.9, 19.2, 19.3, 19.9, 20.1, 19.8, 20.0, 19.7, 19.6, 19.5, 20.2, 20.3, 18.7, 18.8, 17.2, 16.9, 11.5, 11.2, 11.3, 11.4, 7.1, 12.9, 14.4, 13.1, 13.2, 12.5, 12.1, 12.2, 13.0, 0.2, 3.6, 7.4, 4.6, 4.5, 4.3, 4.0, 9.4, 9.6, 9.7, 5.8, 5.7, 7.3, 2.1, 0.4, 0.3, 16.1, 11.9, 12.0, 11.7, 11.8, 10.0, 10.1, 9.8, 15.1, 14.7, 14.8, 10.2, 10.3, 1.2, 9.9, 1.9, 3.4, 14.6, 0.6, 5.1, 5.2, 7.5, 19.4, 10.7, 10.8, 10.9, 0.5, 16.3, 16.2, 16.0, 16.6, 12.4, 11.0, 1.7, 1.6, 2.4, 11.6, 3.9, 3.8, 14.5, 11.1]

count = Counter(map(int, speeds))
print next((s for s in reversed(sorted(speeds))
            if 5 <= s <= 30 and (s <= 14 or count[int(s)] >= 10)),
           False)

您也可以使用Counter：

而不是groupby

speeds = [8.0, 1.3, 0.7, 0.8, 0.9, 1.1, 14.9, 14.0, 14.1, 14.2, 14.3, 13.8, 13.9, 13.7, 13.6, 13.5, 13.4, 15.7, 15.8, 15.0, 15.3, 15.4, 15.5, 15.6, 15.2, 12.8, 12.7, 12.6, 8.7, 8.8, 8.6, 9.0, 8.5, 8.4, 8.3, 0.1, 0.0, 16.4, 16.5, 16.7, 16.8, 17.0, 17.1, 17.8, 17.7, 17.6, 17.4, 17.5, 17.3, 17.9, 18.2, 18.3, 18.1, 18.0, 18.4, 18.5, 18.6, 19.0, 19.1, 18.9, 19.2, 19.3, 19.9, 20.1, 19.8, 20.0, 19.7, 19.6, 19.5, 20.2, 20.3, 18.7, 18.8, 17.2, 16.9, 11.5, 11.2, 11.3, 11.4, 7.1, 12.9, 14.4, 13.1, 13.2, 12.5, 12.1, 12.2, 13.0, 0.2, 3.6, 7.4, 4.6, 4.5, 4.3, 4.0, 9.4, 9.6, 9.7, 5.8, 5.7, 7.3, 2.1, 0.4, 0.3, 16.1, 11.9, 12.0, 11.7, 11.8, 10.0, 10.1, 9.8, 15.1, 14.7, 14.8, 10.2, 10.3, 1.2, 9.9, 1.9, 3.4, 14.6, 0.6, 5.1, 5.2, 7.5, 19.4, 10.7, 10.8, 10.9, 0.5, 16.3, 16.2, 16.0, 16.6, 12.4, 11.0, 1.7, 1.6, 2.4, 11.6, 3.9, 3.8, 14.5, 11.1]

from itertools import *
groups = (list(group) for _, group in groupby(reversed(sorted(speeds)), int))
print next((s[0] for s in groups
            if 5 <= s[0] <= 30 and (s[0] <= 14 or len(s) >= 10)),
           False)

以防所有这些看起来很奇怪，这里有一个接近你原来的。只需查看从最快到最慢的速度并返回符合要求的第一个：

def f(speeds):
    count = Counter(map(int, speeds))
    for speed in reversed(sorted(speeds)):
        if 5 <= speed <= 30 and (speed <= 14 or count[int(speed)] >= 10):
            return speed
    return False

顺便说一句，你对＆＃34;真正的最高速度＆＃34; 的定义对我来说似乎很奇怪。只看某个百分点怎么样？也许是这样的：

print sorted(speeds)[len(speeds) * 9 // 10]

Answer 2

我不确定这是否更快，但它更短，我认为它可以满足您的要求。它使用Counter。

from collections import Counter
import math

def valid(item):
  speed,count = item
  return speed <= 30 and (speed <= 13 or count >= 10)

speeds = [4,3,1,3,4,5,6,7,14,16,18,19,20,34,5,4,3,2,12,58,14,14,14]

speeds = map(math.floor,speeds)
counts = Counter(speeds)
max_valid_speed = max(filter(valid,counts.items()))

结果：max_valid_speed == (12,1)

Answer 3

使用您的排序提示，我们可以在列表末尾以小于30的数字开始，返回匹配条件的第一个数字或返回False：

from collections import Counter

def f(speeds):
    # get speeds that satisfy the range
    rev = [speed for speed in speeds if 5 <= speed < 30]
    rev.sort(reverse=True)
    c = Counter((int(v) for v in rev))
    for speed in rev:
        # will hit highest numbers first
        # so return first that matches
        if speed > 14 and c[int(speed)] > 9 or speed < 15:
            return speed
    # we did not find any speed that matched our requirement
    return False

列表输出v：

In [70]: f(v)
Out[70]: 19.9

如果没有排序，你可以使用dict，根据你的数据是什么决定哪个是最好的，它适用于包括空列表在内的所有情况：

 def f_dict(speeds):
    d = defaultdict(lambda: defaultdict(lambda: 0, {}))
    for speed in speeds:
        key = int(speed)
        d[key]["count"] += 1
        if speed > d[key]["speed"]:
            d[key]["speed"] = speed
    filt = max(filter(lambda x: (15 <= x[0] < 30 and
                                 x[1]["count"] > 9 or x[0] < 15), d.items()), default=False)
    return filt[1]["speed"] if filt else False

输出：

In [95]: f_dict(v)
Out[95]: 19.9

如何提高我的选择过程的速度，python

3 个答案: