我们有一个列表,其元素为字典。我们需要将其展平以得到单个字典,其中所有这些列表元素都以键值对的形式出现。
我们采用一个空字典,并通过从列表中读取元素来向其中添加元素。元素的添加是使用更新功能完成的。
listA = [{'Mon':2}, {'Tue':11}, {'Wed':3}] # printing given arrays print("Given array:\n",listA) print("Type of Object:\n",type(listA)) res = {} for x in listA: res.update(x) # Result print("Flattened object:\n ", res) print("Type of flattened Object:\n",type(res))
输出结果
运行上面的代码给我们以下结果-
('Given array:\n', [{'Mon': 2}, {'Tue': 11}, {'Wed': 3}]) ('Type of Object:\n', ) ('Flattened object:\n ', {'Wed': 3, 'Mon': 2, 'Tue': 11}) ('Type of flattened Object:\n', )
我们还可以将reduce函数与update函数一起使用,以从列表中读取元素并将其添加到空字典中。
from functools import reduce listA = [{'Mon':2}, {'Tue':11}, {'Wed':3}] # printing given arrays print("Given array:\n",listA) print("Type of Object:\n",type(listA)) # Using reduce and update res = reduce(lambda d, src: d.update(src) or d, listA, {}) # Result print("Flattened object:\n ", res) print("Type of flattened Object:\n",type(res))
输出结果
运行上面的代码给我们以下结果-
('Given array:\n', [{'Mon': 2}, {'Tue': 11}, {'Wed': 3}]) ('Type of Object:\n', ) ('Flattened object:\n ', {'Wed': 3, 'Mon': 2, 'Tue': 11}) ('Type of flattened Object:\n', )
ChainMap函数将从列表中读取每个元素,并创建一个新的集合对象,但不创建字典。
from collections import ChainMap listA = [{'Mon':2}, {'Tue':11}, {'Wed':3}] # printing given arrays print("Given array:\n",listA) print("Type of Object:\n",type(listA)) # Using reduce and update res = ChainMap(*listA) # Result print("Flattened object:\n ", res) print("Type of flattened Object:\n",type(res))
输出结果
运行上面的代码给我们以下结果-
Given array: [{'Mon': 2}, {'Tue': 11}, {'Wed': 3}] Type of Object: Flattened object: ChainMap({'Mon': 2}, {'Tue': 11}, {'Wed': 3}) Type of flattened Object: