[英]How to count all elements in a tuple of tuples with dictionary?
I have the following list:我有以下列表:
lst= (1,(1,2), 3, (3,4), 1, 3) lst= (1,(1,2), 3, (3,4), 1, 3)
and I want to use the dictionary function generate output that will count the number of times each value occurs such that it would look like this:我想使用字典 function generate output 来计算每个值出现的次数,它看起来像这样:
{1:3, 2:1, 3:3, 4:1} {1:3, 2:1, 3:3, 4:1}
I am lost on how to do this.我不知道如何做到这一点。 Thank you!
谢谢!
Below is my attempt:以下是我的尝试:
def f(*args):
for x in args:
d = {x:args.count(x) for x in args}
return d
For arbitrary depth tuples, you could use a recursive function for flattening:对于任意深度的元组,您可以使用递归 function 进行展平:
def flatten_nested_tuples(tuples):
for tup in tuples:
if isinstance(tup, tuple):
yield from flatten_nested_tuples(tup)
else:
yield tup
The yield from x
syntax is equivalent to the for item in x: yield item
. yield from x
语法等价于for item in x: yield item
。 Its just a shorter way to create generators.它只是创建生成器的一种更短的方法。 You can have a look at this answer and this answer for more information about generators and the
yield
keyword.您可以查看此答案和此答案,以获取有关生成器和
yield
关键字的更多信息。
To count we can use collections.Counter
to count the flattened tuples:要计数,我们可以使用
collections.Counter
来计算扁平元组:
from collections import Counter
lst= (1,(1,2), 3, (3,4), 1, 3)
print(Counter(flatten_nested_tuples(lst)))
Output: Output:
Counter({1: 3, 3: 3, 2: 1, 4: 1})
Note: Counter
is a subclass of dict
, so you can treat it like a regular dict.注意:
Counter
是dict
的子类,因此您可以将其视为常规 dict。
If you want to count yourself without any modules, you have to do the 0
initializing yourself:如果您想在没有任何模块的情况下计算自己,则必须对自己进行
0
初始化:
counts = {}
for item in flatten_nested_tuples(lst):
counts[item] = counts.get(item, 0) + 1
print(counts)
# {1: 3, 2: 1, 3: 3, 4: 1}
Or without using dict.get()
:或者不使用
dict.get()
:
counts = {}
for item in flatten_nested_tuples(lst):
if item not in counts:
counts[item] = 0
counts[item] += 1
print(counts)
# {1: 3, 2: 1, 3: 3, 4: 1}
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