Question

我正在尝试为下面的元组列表构建一个字典：

lst=[('ldb', 25), ('baseB', 4), ('code', 112), 
     ('cache-6', 55), ('Xauthority', 1), ('baseA', 4), 
     ('npmrc', 1), ('apmrc', 1),('gz', 190), 
     ('dbf', 1), ('lst', 2), ('markdown', 10), 
     ('sqlite-shm', 2), ('vsixmanifest', 4), ('ttf', 109), 
     ('pkl', 35), ('gitignore', 8), ('xml', 46)]

通过使用这样的连接：

op= {','.join( '\'%s\':%d'%i for i in lst)}

但输出op的类型为set，如下所示!!

set(["'ldb':25,'baseB':4,'code':112,'cache-6':55, 'Xauthority':1,'baseA':4,'npmrc':1,'apmrc':1,
'gz':190,'dbf':1,'lst':2,'markdown':10,'sqlite-shm':2,'vsixmanifest':4,'ttf':109,'pkl':35,'gitignore':8,'xml':46"])

有人纠正了我dictionary而不是set 提前谢谢。

Answer 1

目前，您正在创建一个集合，而不是字典。试试这个：

lst=[('ldb', 25), ('baseB', 4), ('code', 112), ('cache-6', 55), ('Xauthority', 1), ('baseA', 4), ('npmrc', 1), ('apmrc', 1), ('gz', 190), ('dbf', 1), ('lst', 2), ('markdown', 10), ('sqlite-shm', 2), ('vsixmanifest', 4), ('ttf', 109), ('pkl', 35), ('gitignore', 8), ('xml', 46)]

new_data = {a:b for a, b in lst}

或者，更好的是：

new_data = dict(lst)

Answer 2

试一试：

lst = [('ldb', 25), ('baseB', 4), ('code', 112), ('cache-6', 55), ('Xauthority', 1), ('baseA', 4), ('npmrc', 1),
   ('apmrc', 1), ('gz', 190), ('dbf', 1), ('lst', 2), ('markdown', 10), ('sqlite-shm', 2), ('vsixmanifest', 4),
   ('ttf', 109), ('pkl', 35), ('gitignore', 8), ('xml', 46)]

d = dict()
for i in lst:
    d[i[0]] = i[1]
print(d)

如何从join构建字典？

2 个答案: