[英]iterate over a dictionary of list of dictionaries in pandas dataframe
How to iterate over a dictionary of list of dictionaries in pandas dataframe one of the column is Response. 如何在pandas数据帧中的字典列表的字典上进行迭代,该列之一是Response。
Response = {"query":"hi","intents":[{"intent":"greeting","score":0.941468239},{"intent":"sentinel","score":7.298465E-06},{"intent":"analyticsPage","score":3.77489937E-06},{"intent":"KaaraServices","score":0.0251461584},{"intent":"goodBye","score":0.357869864},{"intent":"servicesPage","score":2.3839857E-05},{"intent":"PredAnalytics","score":3.21742641E-06},{"intent":"ykaara","score":0.006888155},{"intent":"creator","score":0.061054837}],"entities":[]},
I want to get the intent with the maximum score. 我想获得最高分的意图。
You can turn the value (the list of dicts) of 'intents'
into a DataFrame
and then use argmax
: 您可以将
'intents'
的值( DataFrame
列表)转换为DataFrame
,然后使用argmax
:
df = pd.DataFrame(Response["intents"])
Its output is: 其输出为:
intent score
0 greeting 0.941468
1 sentinel 0.000007
2 analyticsPage 0.000004
3 KaaraServices 0.025146
4 goodBye 0.357870
5 servicesPage 0.000024
6 PredAnalytics 0.000003
7 ykaara 0.006888
8 creator 0.061055
And then for the max: 然后最大:
df.iloc[df['score'].argmax(),:]
this will return: 这将返回:
intent greeting
score 0.941468
Name: 0, dtype: object
I'm sure there's also pandas specific way(s) to do it, but another option is to extract the dictionary and sort it like a generic dictionary. 我敢肯定,还有做熊猫的特定方法,但是另一种选择是提取字典并像普通字典一样对它进行排序。
from operator import itemgetter
# Get pandas frame from somewhere
response = getresponse()
# Convert list of dictionary to dictionary
d = {}
for item in response["intents"]:
d[item["intent"]] = item["score"]
# Sort dictionary, get highest scoring element
sorted(d.items(), key=itemgetter(1), reverse=True)[0]
Try this:- 尝试这个:-
Response = {"query":"hi","intents":[{"intent":"greeting","score":0.941468239},{"intent":"sentinel","score":7.298465E-06},\
{"intent":"analyticsPage","score":3.77489937E-06},{"intent":"KaaraServices","score":0.0251461584},\
{"intent":"goodBye","score":0.357869864},{"intent":"servicesPage","score":2.3839857E-05},\
{"intent":"PredAnalytics","score":3.21742641E-06},{"intent":"ykaara","score":0.006888155},\
{"intent":"creator","score":0.061054837}],"entities":[]},
max_score = []
for i in Response:
for j in i['intents']:
max_score.append(j['score'])
print(max(max_score)) #your expected outpu
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