从整数列表中筛选最多20个值

I'd like to create a list maxValues containing top 20 values from a list of integers lst.

我想从整数列表lst创建一个包含前20个值的列表maxValues。

maxValues = []
for i in range(20):
  maxValues.append(max(lst))
  lst.remove(max(lst))

Is there a more compact code for achieving this task or even built-in function?

是否有更紧凑的代码来实现此任务甚至内置功能?

2 个解决方案

#1

There's heapq.nlargest():

maxvalues = heapq.nlargest(20, lst)

From the doc:

从文档:

heapq.nlargest(n, iterable, key=None)

heapq.nlargest(n,iterable,key = None)

Return a list with the n largest elements from the dataset defined by iterable. key, if provided, specifies a function of one argument that is used to extract a comparison key from each element in the iterable: key=str.lower Equivalent to: sorted(iterable, key=key, reverse=True)[:n]

返回一个列表,其中包含iterable定义的数据集中的n个最大元素。 key(如果提供)指定一个参数的函数,该函数用于从iterable中的每个元素中提取比较键:key = str.lower等效于:sorted(iterable,key = key,reverse = True)[:n]

Or at the same way use heapq.nsmallest() if you want the smallest.

或者以相同的方式使用heapq.nsmallest(),如果你想要最小的。

IMPORTANT NOTE from the doc:

来自doc的重要说明:

The latter two functions [nlargest and nsmallest] perform best for smaller values of n. For larger values, it is more efficient to use the sorted() function. Also, when n==1, it is more efficient to use the built-in min() and max() functions.

后两个函数[nlargest和nsmallest]对较小的n值表现最佳。对于更大的值,使用sorted()函数更有效。此外,当n == 1时,使用内置的min()和max()函数效率更高。

#2

sorted(lst)[-20:]

is the shortest I can think of. Likely to be faster, too.

是我能想到的最短的。也可能更快。

(edited: first try found the min instead of the max)

(编辑:首先尝试找到min而不是max)

#1