对前十个结果进行排序 [英] Sort the top ten results

问题描述

我得到一个列表，其中我以下列方式保存结果

城市百分比孟买 98.30伦敦 23.23阿格拉 12.22.....

列表结构为 [["Mumbai",98.30],["London",23.23]..]

我正在以列表的形式保存这些记录.我需要将列表排序为 top_10 记录.即使我也得到了城市，也可以.

我正在尝试使用以下逻辑，但无法提供准确的数据

if(条件):如果 b 不在 top_ten 中:top_ten.append(b)top_ten.remove(tmp)

也欢迎任何其他解决方案.

编辑 1

 for a in sc_percentage:打印一个

我得到的清单

(, 100.0)(<ServiceCenter: DELHI-DLE>, 75.0)(<服务中心:DELHI-DLN>, 90.909090909090907)(<服务中心:DELHI-DLS>, 83.333333333333343)(<服务中心:DELHI-DLW>, 92.307692307692307)

解决方案

先对列表进行排序再切片:

<预><代码>>>>lis = [['孟买', 98.3], ['伦敦', 23.23], ['阿格拉', 12.22]]>>>print sorted(lis, key = lambda x : x[1], reverse = True)[:10] #[:10] 返回前十项[['孟买'，98.3]，['伦敦'，23.23]，['阿格拉'，12.22]]

要从该文件以列表形式获取数据，请使用:

 with open('abc') as f:next(f) #跳过标题lis = [[city,float(val)] for city, val in(line.split() for line in f)]打印列表#[['孟买', 98.3], ['伦敦', 23.23], ['阿格拉', 12.22]]

更新:

new_lis = sorted(sc_percentage, key = lambda x : x[1], reverse = True)[:10]对于 new_lis 中的项目:打印项目

sorted 返回一个新的排序列表，因为我们需要根据每个元素的第二项对列表进行排序，因此我们使用了 key 参数.

key = lambda x : x[1] 表示使用每个项目的索引 1(即 100.0、75.0 等)上的值进行比较.

reverse= True 用于反向排序.

I am getting a list in which I am saving the results in the following way

City Percentage
Mumbai  98.30
London 23.23
Agra    12.22
.....

List structure is [["Mumbai",98.30],["London",23.23]..]

I am saving this records in form of a list.I need the list to be sort top_ten records.Even if I get cities also, it would be fine.

I am trying to use the following logic, but it fails for to provide accurate data

if (condition):
    if b not in top_ten:
        top_ten.append(b)   
        top_ten.remove(tmp)

Any other solution,approach is also welcome.

EDIT 1

for a in sc_percentage:
            print a

List I am getting

(<ServiceCenter: DELHI-DLC>, 100.0)
(<ServiceCenter: DELHI-DLE>, 75.0)
(<ServiceCenter: DELHI-DLN>, 90.909090909090907)
(<ServiceCenter: DELHI-DLS>, 83.333333333333343)
(<ServiceCenter: DELHI-DLW>, 92.307692307692307)

解决方案

Sort the list first and then slice it:

>>> lis = [['Mumbai', 98.3], ['London', 23.23], ['Agra', 12.22]]
>>> print sorted(lis, key = lambda x : x[1], reverse = True)[:10] #[:10] returns first ten items
[['Mumbai', 98.3], ['London', 23.23], ['Agra', 12.22]]

To get data in list form from that file use this:

with open('abc') as f:
    next(f)  #skip header 
    lis = [[city,float(val)]  for city, val in( line.split() for line in f)]
    print lis 
    #[['Mumbai', 98.3], ['London', 23.23], ['Agra', 12.22]]  

Update:

new_lis = sorted(sc_percentage, key = lambda x : x[1], reverse = True)[:10]
for item in new_lis:
   print item

sorted returns a new sorted list, as we need to sort the list based on the second item of each element so we used the key parameter.

key = lambda x : x[1] means use the value on the index 1(i.e 100.0, 75.0 etc) of each item for comparison.

reverse= True is used for reverse sorting.

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