Question

这不是重复的，因为编写代码的方式使我无法使用我研究堆栈溢出的方法。

我有这样的字典：
```
class_dictionary: {'John':[2,3,4], 'Mark':[6,9,4], 'James':[0,0,0]}
```
我需要打印每个字符从最高到最低的值最高的名称。我已经写过：
```
for names, scores in class_dictionary.items():
        print("{} : {}".format(names, max(scores)))
```
这会打印出每个学生的最高价值但不按顺序排列。像这样：
```
John: 4
James: 0
Mark: 9
```
期望的输出：
```
Mark: 9
John: 4
James: 0
```
我如何将这些结果打印到最低到最低？
我有这样的字典：
```
class_dictionary: {'John':[2,3,4], 'Mark':[6,9,4], 'James':[0,0,0]}
```
我需要打印名称，每个人的平均值从最高到最低。我已经写过：
```
pprint({k: [sum(float(i) for i in v) / len(v)] for k, v in class_dictionary.items()})
```
这打印出每个学生的平均值，但不是按顺序。像这样：
```
John: 3
Mark: 6.33333
James: 0
```
如何将这些结果打印出最高到最低？

期望的输出：
```
Mark: 6.33333
John: 2
James: 0
```

如果您打算投票，请解释原因，以便我下次可以改进问题。

Answer 1

您只需在打印前进行排序：

var MyCtrl = function($route, $scope, $log, $http, $sce, $q, $filter) {

    var vm = this;
    vm.$route = $route;
    vm.$scope = $scope;
    vm.$log = $log;
    vm.$http = $http;
    vm.$sce = $sce;
    vm.$q = $q;
    vm.$filter = $filter;
    vm.activate = activate() ;

    activate()

   function activate() {
        vm.ShouldAutoStart = false;
        vm.viewA = true;
        vm.viewB = false;
        vm.topButtonClicked = false;
        vm.showIDTable = false;
        vm.topRequest;
    }
};

打印：

d = {'John': [2,3,4], 'Mark': [6,9,4], 'James': [0,0,0]}

# Both of the lists below can be generators as well, just change surrounding [] to ()
# Generate a list containing names and highest scores only
highest_scores = [(name, max(scores)) for (name, scores) in d.items()]
# Generate a list containing names and averages only
averages = [(name, sum(scores)/float(len(scores))) for (name, scores) in d.items()]

# Every item is a tuple of name and score, so sort by item[1] (which is the score)
sort_key = lambda item: item[1]

for name, score in sorted(highest_scores, key=sort_key, reverse=True):
    print("%s: %d" % (name, score))

for name, avg in sorted(averages, key=sort_key, reverse=True):
    print("%s: %f" % (name, avg))

Answer 2

将最大值存储在临时字典中，然后再次对其进行排序

class_dictionary= {'Jhon':[2,3,4], 'Mark':[6,9,4], 'James':[0,0,0]}

tmp = {}
for names, scores in class_dictionary.items():
     tmp[names] = max(scores)

print sorted(tmp.items(), reverse=True, key=lambda (k,v) : v )

你明白了。试试你的第二个问题并发布你走了多远，哪些没有用。

Answer 3

import pandas as pd

class_dictionary =  {'Jhon':[2,3,4], 'Mark':[6,9,4], 'James':[0,0,0]}

test_df = pd.DataFrame(class_dictionary)

max_df =  test_df.max().sort_values(ascending = False)
print max_df

mean_df =  test_df.mean().sort_values(ascending = False)
print mean_df

打印字典值从最高到最低

3 个答案: