根据python中的字典值对列表进行排序?

Say I have a dictionary and then I have a list that contains the dictionary's keys. Is there a way to sort the list based off of the dictionaries values?

假设我有一个字典，然后我有一个包含字典键的列表。是否有一种方法可以根据字典的值对列表进行排序?

I have been trying this:

我一直在尝试:

trial_dict = {'*':4, '-':2, '+':3, '/':5}
trial_list = ['-','-','+','/','+','-','*']

I went to use:

我去使用:

sorted(trial_list, key=trial_dict.values())

And got:

和有:

TypeError: 'list' object is not callable

Then I went to go create a function that could be called with trial_dict.get():

然后我去创建一个函数，可以用trial_dict.get()来调用。

def sort_help(x):
    if isinstance(x, dict):
        for i in x:
            return x[i]

sorted(trial_list, key=trial_dict.get(sort_help(trial_dict)))

I don't think the sort_help function is having any affect on the sort though. I'm not sure if using trial_dict.get() is the correct way to go about this either.

我不认为sort_help函数对排序有任何影响。我不确定使用trial_dict.get()是否也是正确的方法。

2 个解决方案

#1

Yes dict.get is the correct (or at least, the simplest) way:

get是正确的(或者至少是最简单的)方法:

sorted(trial_list, key=trial_dict.get)

As Mark Amery commented, the equivalent explicit lambda:

正如Mark Amery所说，等价的显式

sorted(trial_list, key=lambda x: trial_dict[x])

might be better, for at least two reasons:

可能会更好，至少有两个原因:

the sort expression is visible and immediately editable
排序表达式是可见的，并且可以立即编辑
it doesn't suppress errors (when the list contains something that is not in the dict).
它不会抑制错误(当列表中包含不在字典中的内容时)。

#2

The key argument in the sorted builtin function (or the sort method of lists) has to be a function that maps members of the list you're sorting to the values you want to sort by. So you want this:

排序内建函数(或列表的排序方法)中的关键参数必须是一个函数，该函数将正在排序的列表中的成员映射到要排序的值。所以你想要这样的:

sorted(trial_list, key=lambda x: trial_dict[x])

#1