Question

我有一个字典作为全局变量和一个字符串列表：

benchmark.js

我的目标是创建以下列表：

GLOBAL = {"first": "Won't change", "second": ""}
words = ["a", "test"]

我可以使用以下代码执行此操作：

[{"first": "Won't change", "second": "a"}, {"first": "Won't change", "second": "test"}]

我的问题是如何使用理解列表或使用map（）函数

来完成

Answer 1

非常确定你可以在一条丑陋的线条中做到这一点。假设你使用不可变量作为值，否则你必须进行深度复制，这也是可能的：

[GLOBAL.copy().update(second=w) for w in word]

甚至更好（仅限Python 3）

[{**GLOBAL, "second": w} for w in word]

Answer 2

GLOBAL = {"first": "Won't change", "second": ""}
words = ["a", "test"]
result_list = []
for word in words:
    dictionary_to_add = GLOBAL.copy()
    dictionary_to_add["second"] = word
    result_list.append(dictionary_to_add)
print result_list
def hello(word):
    dictionary_to_add = GLOBAL.copy()
    dictionary_to_add["second"] = word
    return dictionary_to_add
print [hello(word) for word in words]
print map(hello,words)

测试它，并尝试更多。

Answer 3

＆＃13;

In [106]: def up(x):
     ...:     d = copy.deepcopy(GLOBAL)
     ...:     d.update(second=x)
     ...:     return d
     ...: 

In [107]: GLOBAL
Out[107]: {'first': "Won't change", 'second': ''}

In [108]: map(up, words)
Out[108]: 
[{'first': "Won't change", 'second': 'a'},
 {'first': "Won't change", 'second': 'test'}]

＆＃13;

使用map或comprehension list python从全局变量创建字典列表

3 个答案: