如何使用python上的循环将字典作为值插入字典
我目前面临的一个问题是如何将我的cvs数据编入字典 我想在文件中使用3列:如何使用python上的循环将字典作为值插入字典,python,loops,dictionary,Python,Loops,Dictionary,我目前面临的一个问题是如何将我的cvs数据编入字典 我想在文件中使用3列: userID, placeID, rating U1000, 12222, 3 U1000, 13333, 2 U1001, 13333, 4 我想让结果如下所示: {'U1000': {'12222': 3, '13333': 2}, 'U1001': {'13333': 4}} 就是说,, 我希望使我的数据结构看起来像: sample = {} sample["U1000"] = {} sam
userID, placeID, rating
U1000, 12222, 3
U1000, 13333, 2
U1001, 13333, 4
我想让结果如下所示:
{'U1000': {'12222': 3, '13333': 2},
'U1001': {'13333': 4}}
就是说,,
我希望使我的数据结构看起来像:
sample = {}
sample["U1000"] = {}
sample["U1001"] = {}
sample["U1000"]["12222"] = 3
sample["U1000"]["13333"] = 2
sample["U1001"]["13333"] = 4
但是我有很多数据要处理。
我想通过循环得到结果,但我已经尝试了2个小时,但失败了
---以下代码可能会使您感到困惑---
现在我的结果是这样的:
{'U1000': ['12222', 3],
'U1001': ['13333', 4]}
reader = np.array(pd.read_csv("rating_final.csv"))
included_cols = [0, 1, 2]
sample= {}
target=[]
target1 =[]
for row in reader:
content = list(row[i] for i in included_cols)
target.append(content[0])
target1.append(content[1:3])
sample = dict(zip(target, target1))
如何改进代码?
我已经看过了,但由于个人能力不足,
谁能帮我一下吗
非常感谢 这应该满足您的要求:
import collections
reader = ...
sample = collections.defaultdict(dict)
for user_id, place_id, rating in reader:
rating = int(rating)
sample[user_id][place_id] = rating
print(sample)
# -> {'U1000': {'12222': 3, '1333': 2}, 'U1001': {'13333': 4}}
是一个方便实用程序,每当您尝试访问字典中不存在的密钥时,它都会提供默认值。如果您不喜欢它(例如,因为您希望sample['non-existence-user-id]
以KeyError
失败),请使用以下方法:
reader = ...
sample = {}
for user_id, place_id, rating in reader:
rating = int(rating)
if user_id not in sample:
sample[user_id] = {}
sample[user_id][place_id] = rating
这应该满足您的要求:
import collections
reader = ...
sample = collections.defaultdict(dict)
for user_id, place_id, rating in reader:
rating = int(rating)
sample[user_id][place_id] = rating
print(sample)
# -> {'U1000': {'12222': 3, '1333': 2}, 'U1001': {'13333': 4}}
是一个方便实用程序,每当您尝试访问字典中不存在的密钥时,它都会提供默认值。如果您不喜欢它(例如,因为您希望sample['non-existence-user-id]
以KeyError
失败),请使用以下方法:
reader = ...
sample = {}
for user_id, place_id, rating in reader:
rating = int(rating)
if user_id not in sample:
sample[user_id] = {}
sample[user_id][place_id] = rating
示例中的预期输出是不可能的,因为
{'1333':2}
不会与键相关联。您可以使用dict的dict
获得{'U1000':{'12222':3,'1333':2},'U1001':{'13333':4}
:
sample = {}
for row in reader:
userID, placeID, rating = row[:3]
sample.setdefault(userID, {})[placeID] = rating # Possibly int(rating)?
或者,使用以避免需要(或涉及try
/的替代方法,除了KeyError
或如果示例中的userID:
牺牲setdefault
的原子性以换取不必要地创建空的dict
):
将其转换回普通dict
可确保将来的查找不会自动激活键,将KeyError
提升为正常,并且如果打印dict
,它看起来像正常的dict
如果包含的列很重要(因为名称或列索引可能会更改),可以使用操作符.itemgetter
一次加速并简化提取所有所需列的过程:
from collections import defaultdict
from operator import itemgetter
included_cols = (0, 1, 2)
# If columns in data were actually:
# rating, foo, bar, userID, placeID
# we'd do this instead, itemgetter will handle all the rest:
# included_cols = (3, 4, 0)
get_cols = itemgetter(*included_cols) # Create function to get needed indices at once
sample = defaultdict(dict)
# map(get_cols, ...) efficiently converts each row to a tuple of just
# the three desired values as it goes, which also lets us unpack directly
# in the for loop, simplifying code even more by naming all variables directly
for userID, placeID, rating in map(get_cols, reader):
sample[userID][placeID] = rating # Possibly int(rating)?
示例中的预期输出是不可能的,因为{'1333':2}
不会与键相关联。您可以使用dict的dict
获得{'U1000':{'12222':3,'1333':2},'U1001':{'13333':4}
:
sample = {}
for row in reader:
userID, placeID, rating = row[:3]
sample.setdefault(userID, {})[placeID] = rating # Possibly int(rating)?
或者,使用以避免需要(或涉及try
/的替代方法,除了KeyError
或如果示例中的userID:
牺牲setdefault
的原子性以换取不必要地创建空的dict
):
将其转换回普通dict
可确保将来的查找不会自动激活键,将KeyError
提升为正常,并且如果打印dict
,它看起来像正常的dict
如果包含的列很重要(因为名称或列索引可能会更改),可以使用操作符.itemgetter
一次加速并简化提取所有所需列的过程:
from collections import defaultdict
from operator import itemgetter
included_cols = (0, 1, 2)
# If columns in data were actually:
# rating, foo, bar, userID, placeID
# we'd do this instead, itemgetter will handle all the rest:
# included_cols = (3, 4, 0)
get_cols = itemgetter(*included_cols) # Create function to get needed indices at once
sample = defaultdict(dict)
# map(get_cols, ...) efficiently converts each row to a tuple of just
# the three desired values as it goes, which also lets us unpack directly
# in the for loop, simplifying code even more by naming all variables directly
for userID, placeID, rating in map(get_cols, reader):
sample[userID][placeID] = rating # Possibly int(rating)?
这似乎是希望字典作为值,而不是键。也许可以更正标题以匹配?谢谢您的提醒。已经更正了标题和内容!另外,您的示例有{'U1000':{'12222':3},{'1333':2},'U1001':{'13333':4}
,但这是映射U1000
和U1001
,但没有与{'1333':2}
关联的键(或没有值)。你可以有{'U1000':{'12222':3,'1333':2},'U1001':{'13333':4}
,或者{'U1000':[{'12222':3},{'1333':2},'U1001':[{'13333':4}
,但不是你提供的。我明白你的意思。谢谢你的澄清。我已经提前纠正了身体。谢谢这似乎是希望字典作为值,而不是键。也许可以更正标题以匹配?谢谢您的提醒。已经更正了标题和内容!另外,您的示例有{'U1000':{'12222':3},{'1333':2},'U1001':{'13333':4}
,但这是映射U1000
和U1001
,但没有与{'1333':2}
关联的键(或没有值)。你可以有{'U1000':{'12222':3,'1333':2},'U1001':{'13333':4}
,或者{'U1000':[{'12222':3},{'1333':2},'U1001':[{'13333':4}
,但不是你提供的。我明白你的意思。谢谢你的澄清。我已经提前纠正了身体。谢谢谢谢你的澄清,这真的很有帮助!谢谢你的澄清,这真的很有帮助!谢谢你的回答,这真的很有帮助!谢谢你的回答,这真的很有帮助!