[英]How to create a nested dictionary in python with 6 lists
我希望扩展此处采用的方法,但对于六个或更多列表的情况: How to Create Nested Dictionary in Python with 3 lists
a = ['A', 'B', 'C', 'D']
b = [1, 2, 3, 4]
c = [9, 8, 7, 6]
d = [0, 3, 5, 7]
e = [11, 13, 14, 15]
所需的 output:
{'A':{1 :9, 0:11} , 'B':{2:8, 3:13}, 'C':{3:7, 5:13} , 'D':{4:6, 7:15}}
到目前为止,这是我尝试过的:
out = dict([[a, dict([map(str, i)])] for a, i in zip(a, zip(zip(b, c), zip(d,e) ))])
output 很接近,但它不是我要找的。 任何提示将非常感谢!
也许是这样的:
out = {k: {k1: v1, k2: v2} for k, k1, v1, k2, v2 in zip(a, b, c, d, e)}
如果我们想使用过多的zip
s,我们也可以这样做:
out = {k: dict(v) for k,v in zip(a, zip(zip(b, c), zip(d, e)))}
Output:
{'A':{1 :9, 0:11} , 'B':{2:8, 3:13}, 'C':{3:7, 5:13} , 'D':{4:6, 7:15}}
这是一种使用两个字典理解和zip()
的方法:
result = {key: { inner_key: inner_value for inner_key, inner_value in value}
for key, value in zip(a, zip(zip(b, c), zip(d, e)))
}
print(result)
结果,使用原始问题中的列表:
{'A': {1: 9, 0: 11}, 'B': {2: 8, 3: 13}, 'C': {3: 7, 5: 14}, 'D': {4: 6, 7: 15}}
你可以试试NestedDict
。 首先安装ndicts
pip install ndicts
然后
from ndicts.ndicts import NestedDict
a = ['A', 'B', 'C', 'D']
b = [1, 2, 3, 4]
c = [9, 8, 7, 6]
d = [0, 3, 5, 7]
e = [11, 13, 14, 15]
# Create levels and values
level_0 = a * 2
level_1 = b + d
values = c + e
# Initialize the keys of a NestedDict
nd = NestedDict.from_tuples(*zip(level_0, level_1))
# Assign values
for key, value in zip(nd, values):
nd[key] = value
如果您需要将结果作为字典
result = nd.to_dict()
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