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[英]Create a list of dictionaries from a list of keys and multiple lists of values
[英]Python 3: Creating list of multiple dictionaries have same keys but different values coming from multiple lists
我正在使用 lxml 库中的 xpath 解析 XML 的响应。 我得到结果并从中创建列表,如下所示:
object_name = [o.text for o in response.xpath('//*[name()="objectName"]')]
object_size_KB = [o.text for o in response.xpath('//*[name()="objectSize"]')]
我想使用列表为列表中的每个元素创建一个字典,然后将它们添加到最终列表中,如下所示:
[{'object_name': 'file1234', 'object_size_KB': 9347627},
{'object_name': 'file5671', 'objeobject_size_KBt_size': 9406875}]
我想要一个生成器,因为我将来可能需要从响应中搜索更多元数据,所以我希望我的代码能够面向未来并减少重复:
meta_names = {
'object_name': '//*[name()="objectName"]',
'object_size_KB': '//*[name()="objectSize"]'
}
def parse_response(response, meta_names):
"""
input: response: api xml response text from lxml xpath
input: meta_names: key names used to generate dictionary per object
return: list of objects dictionary
"""
mylist = []
# create list of each xpath match assign them to variables
for key, value in meta_names.items():
mylist.append({key: [o.text for o in response.xpath(value)]})
return mylist
然而 function 给了我这个:
[{'object_name': ['file1234', 'file5671']}, {'object_size_KB': ['9347627', '9406875']}]
我一直在论坛中寻找类似的案例,但找不到符合我需求的东西。 感谢你的帮助。
更新: Renneys 的答案是我想要的我刚刚调整了结果范围的长度值,因为我并不总是每个 object 键的 xpath 的长度相同,而且每次我选择第一个索引 [0] 时我的列表都有相同的长度。 现在 function 看起来像这样。
def create_entries(root, keys):
tmp = []
for key in keys:
tmp.append([o.text for o in root.xpath('//*[name()="' + key + '"]')])
ret = []
# print(len(tmp[0]))
for i in range(len(tmp[0])):
add = {}
for j in range(len(keys)):
add[keys[j]] = tmp[j][i]
ret.append(add)
return ret
我想这就是你要找的。
您可以使用 zip 将您的两个列表组合成一个值对列表。
然后,您可以使用列表推导式或生成器表达式将值对与所需的键配对。
import pprint
object_name = ['file1234', 'file5671']
object_size = [9347627, 9406875]
[{'object_name': 'file1234', 'object_size_KB': 9347627},
{'object_name': 'file5671', 'objeobject_size_KBt_size': 9406875}]
[{'object_name': ['file1234', 'file5671']}, {'object_size_KB': ['9347627', '9406875']}]
# List Comprehension
obj_list = [{'object_name': name, 'object_size': size} for name,size in zip(object_name,object_size)]
pprint.pprint(obj_list)
print('\n')
# Generator Expression
generator = ({'object_name': name, 'object_size': size} for name,size in zip(object_name,object_size))
for obj in generator:
print(obj)
实时代码示例 -> https://onlinegdb.com/SyNSwd7jU
我认为接受的答案更有效,但这里有一个如何使用列表推导的示例。
meta_names = {
'object_name': ['file1234', 'file5671'],
'object_size_KB': ['9347627', '9406875'],
'object_text': ['Bob', 'Ross']
}
def parse_response(meta_names):
"""
input: response: api xml response text from lxml xpath
input: meta_names: key names used to generate dictionary per object
return: list of objects dictionary
"""
# List comprehensions
to_dict = lambda l: [{key:val for key,val in pairs} for pairs in l]
objs = list(zip(*list([[key,val] for val in vals] for key,vals in meta_names.items())))
pprint.pprint(to_dict(objs))
parse_response(meta_names)
使用二维数组:
def createEntries(root, keys):
tmp = []
for key in keys:
tmp.append([o.text for o in root.xpath('//*[name()="' + key + '"]')])
ret = []
for i in range(len(tmp)):
add = {}
for j in range(len(keys)):
add[keys[j]] = tmp[j][i]
ret.append(add)
return ret
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