用列表理解修改词典列表

使用列表理解和嵌套的dict理解:

new_list = [{ k: v * 2 if k == 'two' else v for k,v in x.items()} for x in myList]
print (new_list)
[{'one': 1, 'two': 4, 'three': 3}, 
 {'one': 4, 'two': 10, 'three': 6}, 
 {'one': 7, 'two': 16, 'three': 9}]

myList = [ {'one':1, 'two':2,'three':3},{'one':4, 'two':5,'three':6},{'one':7, 'two':8,'three':9}  ]

[ { k: 2*i[k] if k == 'two' else i[k] for k in i } for i in myList ]

[{'one': 1, 'three': 3, 'two': 4}, {'one': 4, 'three': 6, 'two': 10}, {'one': 7, 'three': 9, 'two': 16}]

简单的 for 循环应该足够了。但是,如果您想使用字典理解,我发现定义映射字典比三元语句更具可读性和可扩展性:

factor = {'two': 2}

res = [{k: v*factor.get(k, 1) for k, v in d.items()} for d in myList]

print(res)

[{'one': 1, 'two': 4, 'three': 3},
 {'one': 4, 'two': 10, 'three': 6},
 {'one': 7, 'two': 16, 'three': 9}]

你好你试过这个了吗:

for d in myList:
  d.update((k, v*2) for k, v in d.iteritems() if k == "two")

谢谢

在Python3.5+中,可以在 PEP 448 是的。这将创建每个dict的副本,然后覆盖密钥的值 two 以下内容:

new_list = [{**d, 'two': d['two']*2} for d in myList]
# result:
# [{'one': 1, 'two': 4, 'three': 3},
#  {'one': 4, 'two': 10, 'three': 6},
#  {'one': 7, 'two': 16, 'three': 9}]