3 回答
TA贡献1803条经验 获得超3个赞
尝试这个,
key, value = sorted(Dict_1.items(), key = lambda x : x[1], reverse=True)[0]
print(f"{key}, {list_1.count(key) * value}")
# apples, 10.46
TA贡献1818条经验 获得超11个赞
您可以使用它Counter来计算列表中每个项目的频率(出现次数)。
max(counter.values())会给我们最大出现元素的计数
max(counter, key=counter.get)将给出列表中的哪个项目与最高计数相关联。
========================================================================
from collections import Counter
def H_abs_W(list_1, Dict_1):
counter = Counter(list_1)
count = max(counter.values())
item = max(counter, key=counter.get)
return item, abs(count * Dict_1.get(item))
TA贡献1993条经验 获得超6个赞
list_1 = ['apples','oranges','pears','apples']
Dict_1 = {'apples':5.23,'pears':-7.62}
d = {k:0 for k in list_1}
for x in list_1:
if x in Dict_1.keys():
d[x]+=Dict_1[x]
m = max(Dict_1, key=Dict_1.get)
print(m,Dict_1[m])
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