Python在组合字典的列表中删除重复值 [英] Python remove duplicate value in a combined dictionary's list
问题描述
f = {'a':'apple','c' 'cat','b':'bat','d':'dog'}
g = {'c':'car','b':'bat','e':'elephant'}
h = {'b':'boy','d':'deer'}
r = {'a':'adam'}
def merge(* d) :
newdicts = {}
for d中的dict:
在dict.items()中的k:
如果k [0]在newdicts中:
newdicts [k [0]]。append(k [1])$ b $ b else:
newdicts [k [0]] = [k [1]]
return newdicts
组合=合并(f,g,h,r)
打印(合并)
输出如下:
{'a':['apple','adam'],'c':['cat','car'],'b ''['蝙蝠','蝙蝠','男孩'],'e':['elephant'],'d':['dog','deer']}
在b键下,bat出现两次。我如何删除重复的文件?
我已经看过滤镜,lambda,但我无法弄清楚如何使用(也许b / c它是一个列表一个字典?)
任何帮助将不胜感激。谢谢你所有的帮助!
在添加列表之前,测试列表中的元素:如果k(0)在newdicts中,则
1]不在newdicts [k [0]]:#在添加前做这个测试。
newdicts [k [0]]。append(k [1])$ b $ b else:
newdicts [k [0]] = [k [1]]
由于您只想在值
列表,那么您可以使用 Set
作为值。此外,您可以在此处使用 defaultdict
,以便在添加之前不必对密钥存在进行测试。
此外,不要使用内置的作为您的变量名。而不是 dict
其他一些变量。
所以,你可以修改你的 code>方法为:
从集合导入defaultdict
def merge(* d) :
newdicts = defaultdict(set)#在d中为each_dict定义一个defaultdict
#dict.items()返回一个(k,v)元组的列表。
#所以,你可以在两个循环变量中直接解包元组。
for k,v in each_dict.items():
newdicts [k] .add(v)
#如果你想要确切的表示,你已经显示
#你可以用新建的dict命令创建一个正常的命令。
unique = {key:list(value)为key,value in newdicts.items()}
返回唯一
I need a little bit of homework help. I have to write a function that combines several dictionaries into new dictionary. If a key appears more than once; the values corresponding to that key in the new dictionary should be a unique list. As an example this is what I have so far:
f = {'a': 'apple', 'c': 'cat', 'b': 'bat', 'd': 'dog'}
g = {'c': 'car', 'b': 'bat', 'e': 'elephant'}
h = {'b': 'boy', 'd': 'deer'}
r = {'a': 'adam'}
def merge(*d):
newdicts={}
for dict in d:
for k in dict.items():
if k[0] in newdicts:
newdicts[k[0]].append(k[1])
else:
newdicts[k[0]]=[k[1]]
return newdicts
combined = merge(f, g, h, r)
print(combined)
The output looks like:
{'a': ['apple', 'adam'], 'c': ['cat', 'car'], 'b': ['bat', 'bat', 'boy'], 'e': ['elephant'], 'd': ['dog', 'deer']}
Under the 'b' key, 'bat' appears twice. How do I remove the duplicates?
I've looked under filter, lambda but I couldn't figure out how to use with (maybe b/c it's a list in a dictionary?)
Any help would be appreciated. And thank you in advance for all your help!
Just test for the element inside the list before adding it: -
for k in dict.items():
if k[0] in newdicts:
if k[1] not in newdicts[k[0]]: # Do this test before adding.
newdicts[k[0]].append(k[1])
else:
newdicts[k[0]]=[k[1]]
And since you want just unique elements in the value
list, then you can just use a Set
as value instead. Also, you can use a defaultdict
here, so that you don't have to test for key existence before adding.
Also, don't use built-in for your as your variable names. Instead of dict
some other variable.
So, you can modify your merge
method as:
from collections import defaultdict
def merge(*d):
newdicts = defaultdict(set) # Define a defaultdict
for each_dict in d:
# dict.items() returns a list of (k, v) tuple.
# So, you can directly unpack the tuple in two loop variables.
for k, v in each_dict.items():
newdicts[k].add(v)
# And if you want the exact representation that you have shown
# You can build a normal dict out of your newly built dict.
unique = {key: list(value) for key, value in newdicts.items()}
return unique
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