如果字典中的值与列表中的元素匹配,我希望返回给定字典的所有键。假设我有以下字典:
my_dict = {'flower1': ['blue'],
'flower2': ['red', 'green', 'blue'],
'flower3': ['yellow'],
'flower4': ['blue', 'black', 'cyan']}现在,我希望将字典中的值与列表中的下列元素相匹配:
my_lst = ['black',
'red',
'blue',
'yellow',
'green',
'purple',
'brown',
'cyan']我的目标是得到一本字典,如下所示:
result_dict = {'black': ['flower4'],
'red': ['flower2'],
'blue': ['flower1', 'flower2', 'flower4'],
'yellow': ['flower3']
'green': ['flower2'],
'purple': [],
'brown': [],
'cyan': []}现在,我尝试了一个简单的列表理解,它工作得很好,但是只返回一个简单且无序的列表,如下所示:
In[14]: [key for key, value in my_dict.items() for i in range(0, len(my_lst)) if my_lst[i] in value]
Out[14]:['flower1',
'flower2',
'flower2',
'flower2',
'flower3',
'flower4',
'flower4',
'flower4']做这样的手术最好的方法是什么?我不能把头挪开,任何帮助都会很感激的。
发布于 2021-02-12 11:54:15
别把事情弄得太复杂。分两个完全独立的步骤:
my_lst转换为合适的result result,并将它们添加到result字典中。例如:
# create result dict, adding each color as a key
# make the value of each key an empty list initially
result = {k: [] for k in my_lst}
# iterate through the items of the flower/color dict
for flower, colors in my_dict.items():
# append the flower corresponding to each color
# to the appropriate list in the result dict
for color in colors:
result[color].append(flower)
print(result)输出:
{'black': ['flower4'], 'red': ['flower2'], 'blue': ['flower1', 'flower2', 'flower4'], 'yellow': ['flower3'], 'green': ['flower2'], 'purple': [], 'brown': [], 'cyan': ['flower4']}当然,这假设my_dict中的每种颜色都发生在my_lst中。
发布于 2021-02-12 11:54:53
这可以使用两个for循环来实现。它的可读性远比使用一条线。我提供了一个代码片段,它完全按照您的要求执行,理解它的工作原理是相当直观的。您接受每个list元素,并检查每个键是否可以在值列表中找到该元素。
result_dict = {}
for ele in my_lst:
result_dict[ele] = []
for key in my_dict.keys():
if ele in my_dict[key]:
result_dict[ele].append(key)发布于 2021-02-12 11:55:32
此函数可用于替换您的“result_dict”:
def flowers_by_color(color, data = my_dict):
return [flower for flower in data.keys() if color in data[flower]]
result_dict = {color:flowers_by_color(color) for color in my_lst}https://stackoverflow.com/questions/66171273
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