[英]Python: count specific occurrences in a dictionary
Say I have a dictionary like this: 假设我有一本这样的字典:
d={
'0101001':(1,0.0),
'0101002':(2,0.0),
'0101003':(3,0.5),
'0103001':(1,0.0),
'0103002':(2,0.9),
'0103003':(3,0.4),
'0105001':(1,0.0),
'0105002':(2,1.0),
'0105003':(3,0.0)}
Considering that the first four digits of each key consitute the identifier of a "slot" of elements (eg, '0101', '0103', '0105'), how can I count the number of occurrences of 0.0
for each slot? 考虑到每个键的前四位数字构成元素“槽”的标识符(例如,“ 0101”,“ 0103”,“ 0105”),我如何计算每个槽的出现次数
0.0
?
The intended outcome is a dict like this: 预期的结果是这样的字典:
result={
'0101': 2,
'0103': 1,
'0105': 2}
Apologies for not being able to provide my attempt as I don't really know how to do this. 很抱歉无法提供我的尝试,因为我真的不知道该怎么做。
You can use a defaultdict
: 您可以使用
defaultdict
:
from _collections import defaultdict
res = defaultdict(int)
for k in d:
if d[k][1] == 0.0:
res[k[:4]] += 1
print(dict(res))
When you do the +=1
, if the key does not exist, it creates it with value 0
and then does the operation. 当您执行
+=1
,如果键不存在,它将创建一个值为0
,然后执行该操作。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.