python按值的长度排序字典
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python sorting dictionary by length of values
提问by jwillis0720
I have found many threads for sorting by values like herebut it doesn't seem to be working for me...
我发现很多线程可以像这里这样按值排序，但它似乎对我不起作用......
I have a dictionary of lists that have tuples. Each list has a different amount of tuples. I want to sort the dictionary by how many tuples each list contain.
我有一个包含元组的列表字典。每个列表都有不同数量的元组。我想按每个列表包含多少元组对字典进行排序。
>>>to_format
>>>{"one":[(1,3),(1,4)],"two":[(1,2),(1,2),(1,3)],"three":[(1,1)]}
>>>for key in some_sort(to_format):
print key,
>>>two one three
Is this possible?
这可能吗？
采纳答案by jamylak
>>> d = {"one": [(1,3),(1,4)], "two": [(1,2),(1,2),(1,3)], "three": [(1,1)]}
>>> for k in sorted(d, key=lambda k: len(d[k]), reverse=True):
print k,
two one three
Here is a universal solution that works on Python 2 & Python 3:
这是适用于 Python 2 和 Python 3 的通用解决方案：
>>> print(' '.join(sorted(d, key=lambda k: len(d[k]), reverse=True)))
two one three
回答by Oleksiy Tsebriy
dict= {'a': [9,2,3,4,5], 'b': [1,2,3,4, 5, 6], 'c': [], 'd': [1,2,3,4], 'e': [1,2]}
dict_temp = {'a': 'hello', 'b': 'bye', 'c': '', 'd': 'aa', 'e': 'zz'}
def sort_by_values_len(dict):
dict_len= {key: len(value) for key, value in dict.items()}
import operator
sorted_key_list = sorted(dict_len.items(), key=operator.itemgetter(1), reverse=True)
sorted_dict = [{item[0]: dict[item [0]]} for item in sorted_key_list]
return sorted_dict
print (sort_by_values_len(dict))
output:
[{'b': [1, 2, 3, 4, 5, 6]}, {'a': [9, 2, 3, 4, 5]}, {'d': [1, 2, 3, 4]}, {'e': [1, 2]}, {'c': []}]