def sortedDictValues2(adict): keys = adict.keys() keys.sort() return [dict[key] for key in keys] 1. 2. 3. 4. 还是按 key 值排序,据说更快。。。而且当 key 为 tuple 的时候照样适用 def sortedDictValues3(adict): keys = adict.keys() keys.sort() return map(adict.get, keys) ...
items.sort()return[valueforkey, valueinitems] defsortedDictValues2(adict): keys = adict.keys() keys.sort()return[dict[key]forkeyinkeys] defsortedDictValues3(adict): keys = adict.keys() keys.sort()returnmap(adict.get, keys) #一行语句搞定:[(k,di[k])forkinsorted(di.keys())] 按...
items.sort() return [value for key, value in items] #又一个按照key值排序,貌似比上一个速度要快点 def sortedDictValues2(adict): keys = adict.keys() keys.sort() return [dict[key] for key in keys] #还是按key值排序,据说更快。。。而且当key为tuple的时候照样适用 def sortedDictValues3(...
defsort_dict(a_dict,option="value"):'''对dict进行排序:param a_dict: 待排序的字典:param option: 有两种选择,一种是value代表根据value进行排序,一种是key代表根据key值进行排序:return: 排序后的新字典'''ifoptionin["value","key"]:result_dict={}ifoption=="key":temp_list=list(a_dict.keys()...
items.sort()return[valueforkey, valueinitems] 又一个按照key值排序,貌似比上一个速度要快点 defsortedDictValues2(adict): keys = adict.keys() keys.sort()return[dict[key]forkeyinkeys] 还是按key值排序,据说更快。。。而且当key为tuple的时候照样适用 ...
return [dict[key] for key in keys] #还是按key值排序,据说更快。。。而且当key为tuple的时候照样适用 def sortedDictValues3(adict): keys = adict.keys() keys.sort() return map(adict.get, keys) #一行语句搞定: [(k,di[k]) for k in sorted(di.keys())] ...
方法1:最简单的方法,排列元素(key/value对),然后挑出值。字典的items方法,会返回一个元组的列表,其中每个元组都包含一对项目 ——键与对应的值。此时排序可以sort()方法。 def sortedDictValues1(adict): items = adict.items() items.sort() return [value ...
1. sorted是python的内置函数,可以对列表(list),元祖(tuple),字典(dict)和字符串(str)进行排序,排序对象作为sorted函数的参数,使用示例如下: a_tuple=(1,3,2,4)sorted(a_list)(1,2,3,4)#返回 2. sort() 是列表类的方法,只能对列表排序。sorted()对列表排序时,有返回值;sorte()对列表排序时,无法返回...
A simple way I found to sort a dictionary is to create a new one, based on the sorted key:value items of the one you're trying to sort. If you want to sort dict = {}, retrieve all its items using the associated method, sort them using the sorted() function then create the new...
As simple as: sorted(dict1, key=dict1.get) Well, it is actually possible to do a "sort by dictionary values". Recently I had to do that in a Code Golf (Stack Overflow question Code golf: Word frequency chart). Abridged, the problem was of the kind: given a text, count how often...