我有一個由 4 個鍵組成的字典，每個鍵有 3 個值?？雌饋硐襁@樣： d = {key1: (value1, value2, value), key2: (value1, value2, value), key3: (value1, value 2, value3)} 我想從所有鍵打印 value1 。我現在的做法是這樣的：print (persons['1'][0])print(persons['2'][0])print (persons['3'][0])print(persons['4'][0])但我想有一種更簡單的方法可以讓我在一行中引用所有鍵？我還想找到所有鍵中的最高值2，以及所有鍵中值3的平均值。有人可以幫我解決這個問題嗎？
5 回答
有只小跳蛙
TA貢獻1824條經驗 獲得超8個贊
您可以將您的字典轉換為 DataFrame，這將使您的事情變得非常簡單：
from pandas.DataFrame import from_dict
d = {'a':(1,2,3),'b':(4,5,6)}
d = from_dict(d, orient='index')
d[0] # print values of value1
d[1].max() # max of value2
d[2].mean() # mean of value3
qq_花開花謝_0
TA貢獻1835條經驗 獲得超7個贊
您可以使用列表理解來實現此目的：
persons = {'1': (1,2,3), '2': (4,5,6), '3': (7,8,9)}
# First Value's
first_values = " ".join([str(x[0]) for x in persons.values()])
print(first_values) # prints 1 4 7
# Max of second value's
max_value2 = max([x[1] for x in persons.values()])
print(max_value2) # prints 8
# Average of value3's
third_values = [x[2] for x in persons.values()]
average_of_third_values = sum(third_values) / len(third_values)
# in case avoid zero division :
# average_of_third_values = sum(third_values) / (len(third_values) or 1)
print(average_of_third_values) # prints 6
# to get value1 of values which has max value2
value1_of_max = [x[0] for x in persons.values() if x[1]==max_value2]
print(value1_of_max) # prints [7]
# Its possible to be exist more than 1 person that has value2 which equals to max number, like so
# persons = {'1': (1,2,3), '2': (4,8,6), '3': (7,8,9)}
# so we print them as list
PIPIONE
TA貢獻1829條經驗 獲得超9個贊
關于什么？
d = {
"key1": (1,2,3),
"key2": (3,4,5),
"key4": (5,6,8),
}
[ print(val[0]) for _, val in d.items()]
UYOU
TA貢獻1878條經驗 獲得超4個贊
關于什么？
d = {
"key1": (1,2,3),
"key2": (3,4,5),
"key4": (5,6,8),
}
[ print(val[0]) for _, val in d.items()]
縹緲止盈
TA貢獻2041條經驗 獲得超4個贊
您可以使用for loop迭代：
d = {'test1': (1, 2, 3), 'test2': (4, 5, 6), 'test3': (7, 8, 9)}
for key in d.values():
print(key[0])
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