根据另一个列表的值顺序对字典列表进行排序 [英] Sorting a list of dictionaries based on the order of values of another list

问题描述

我正在使用python 2.7.3，并且试图根据另一个列表的值顺序对字典列表进行排序.

I'm using python 2.7.3, and I'm trying to sort a list of dictionaries based on the order of values of another list.

IE:

listOne = ['hazel', 'blue', 'green', 'brown']
listTwo = [{'name': 'Steve', 'eyecolor': 'hazel', 'height': '5 ft. 11 inches'},
           {'name': 'Mark', 'eyecolor': 'brown', 'height': '6 ft. 2 inches'},
           {'name': 'Mike', 'eyecolor': 'blue', 'height': '6 ft. 0 inches'},
           {'name': 'Ryan', 'eyecolor': 'brown', 'height': '6 ft, 0 inches'},
           {'name': 'Amy', 'eyecolor': 'green', 'height': '5 ft, 6 inches'}]

根据listOne中值的顺序对listTwo进行排序，我们将得到以下结果:

Sorting listTwo based off of the order of values in listOne, we would end up with the following:

print listTwo
[{'name': 'Steve', 'eyecolor': 'hazel', 'height': '5 ft. 11 inches'},
{'name': 'Mike', 'eyecolor': 'blue', 'height': '6 ft. 0 inches'},
{'name': 'Amy', 'eyecolor': 'green', 'height': '5 ft, 6 inches'},
{'name': 'Mark', 'eyecolor': 'brown', 'height': '6 ft. 2 inches'},
{'name': 'Ryan', 'eyecolor': 'brown', 'height': '6 ft, 0 inches'}]

我最终需要输出此文本，因此我为正确显示(以正确的顺序)所做的事情如下:

I eventually need to output this text, so what I've done to display it correctly (in the correct order) is the following:

for x in xrange(len(listOne)):
    for y in xrange(len(listTwo)):
        if listOne[x] == listTwo[y]["eyecolor"]:
            print "Name: " + str(listTwo[y]["name"]),
            print "Eye Color: " + str(listTwo[y]["eyecolor"]),
            print "Height: " + str(listTwo[y]["height"])

是否存在某种可用于实现此目的的lambda表达式?必须有一种更紧凑，更简单的方法来按我想要的顺序来获取它.

Is there some sort of lambda expression that can be used to make this happen? There has to be a more compact, less complex way of getting it in the order I want.

推荐答案

最简单的方法是使用list.index为词典列表生成排序值:

The simplest way would be to use list.index to generate a sort value for your list of dictionaries:

listTwo.sort(key=lambda x: listOne.index(x["eyecolor"]))

但这有点效率低下，因为list.index在眼睛颜色列表中进行了线性搜索.如果您要检查的眼睛颜色很多，那将会很慢.更好的方法是改为建立索引字典:

This is a little bit inefficient though, since list.index does a linear search through the eye-color list. If you had many eye colors to check against, it would be slow. A somewhat better approach would build an index dictionary instead:

order_dict = {color: index for index, color in enumerate(listOne)}
listTwo.sort(key=lambda x: order_dict[x["eyecolor"]])

如果不想修改listTwo，则可以使用内置的sorted函数而不是list.sort方法.它返回列表的排序副本，而不是就地排序.

If you don't want to modify listTwo, you can use the built-in sorted function instead of the list.sort method. It returns a sorted copy of the list, rather than sorting in-place.

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