如何用(python)列表中的顺序替换数字 [英] How to replace numbers with order in (python) list

问题描述

我有一个包含整数的列表，想要替换它们，以便以前包含最高数字的元素现在包含1，第二高数字设置为2，依此类推.

I have a list containing integers and want to replace them so that the element which previously contained the highest number now contains a 1, the second highest number set to 2, etc etc.

示例: [5, 6, 34, 1, 9, 3]应该产生[4, 3, 1, 6, 2, 5].

我个人只关心前9个最高的数字，因为我认为可能有一个简单的算法甚至一个python函数来完成这项任务?

I personally only care about the first 9 highest numbers by I thought there might be a simple algorithm or possibly even a python function to do take care of this task?

我不在乎如何处理重复项.

I don't care how duplicates are handled.

推荐答案

一种快速的方法是首先生成该元素的元组列表及其位置:

A fast way to do this is to first generate a list of tuples of the element and its position:

sort_data = [(x,i) for i,x in enumerate(data)]

接下来，我们在reverse中对这些元素进行排序:

next we sort these elements in reverse:

sort_data = sorted(sort_data,reverse=True)

生成的(用于您的示例输入):

which generates (for your sample input):

>>> sort_data
[(34, 2), (9, 4), (6, 1), (5, 0), (3, 5), (1, 3)]

并嵌套，我们需要填写以下元素:

and nest we need to fill in these elements like:

result = [0]*len(data)
for i,(_,idx) in enumerate(sort_data,1):
    result[idx] = i

或将它们放在一起:

def obtain_rank(data):
    sort_data = [(x,i) for i,x in enumerate(data)]
    sort_data = sorted(sort_data,reverse=True)
    result = [0]*len(data)
    for i,(_,idx) in enumerate(sort_data,1):
        result[idx] = i
    return result

此方法适用于 O(n log n)，其中 n 为data中的元素数量.

this approach works in O(n log n) with n the number of elements in data.

更紧凑的算法(在某种意义上说，没有用于排序的元组构成)是

A more compact algorithm (in the sense that no tuples are constructed for the sorting) is:

def obtain_rank(data):
    sort_data = sorted(range(len(data)),key=lambda i:data[i],reverse=True)
    result = [0]*len(data)
    for i,idx in enumerate(sort_data,1):
        result[idx] = i
    return result

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