如果字典中的值与列表的元素匹配,我想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']}
现在我想将字典中的值与列表中的以下元素进行匹配:
my_lst = ['black',
'red',
'blue',
'yellow',
'green',
'purple',
'brown',
'cyan']
我的目标是得到一个如下的字典:
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,它工作正常,但只返回一个简单的无序列表,如:
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']
执行此类操作的最佳方法是什么?我无法理解它,任何帮助将不胜感激。
答案 0 :(得分:2)
不要过度复杂化。分两步完成:
my_lst
转换为合适的 result
字典。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
中的每种颜色都出现在 my_lst
中。
答案 1 :(得分:2)
这可以使用两个 for 循环来实现。它比使用单行更具可读性。我提供了一个代码片段,它完全符合您的要求,理解它的工作原理是相当直观的。您获取每个列表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)
答案 2 :(得分:2)
这个函数可以用来代替你的“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}
答案 3 :(得分:1)
你可以这样做:
result_dict = {color: [] for color in my_lst}
for flower in my_dict:
for color in my_dict[flower]:
if color in my_lst:
result_dict[color].append(flower)