Question

我可以用两种方式写出追加线。都没有产生预期的结果。有没有办法将它包装成一行？

选项1：

row.append(x[1] for x in emails if x[0] == row[1])

收率：

[['project1', 'email1', <generator object <genexpr> at 0x0227D670>], ['project1', 'email2', <generator object <genexpr> at 0x022EB8A0>]]

选项2：

row.append([x[1] for x in emails if x[0] == row[1]])

收率：

[['project1', 'email1', ['john@gmail.com']], ['project1', 'email2', ['bill@gmail.com']]]

期望的结果：

[['project1', 'email1', 'john@gmail.com'], ['project1', 'email2', 'bill@gmail.com']]

代码：

emails = [['email1','john@gmail.com'],['email2','bill@gmail.com']]

projects = [['project1', 'email1'], ['project1', 'email2']]

for row in projects:
    row.append(x[1] for x in emails if x[0] == row[1])
print projects

Answer 1

在现有代码中，替换以下一行：

row.append(x[1] for x in emails if x[0] == row[1])

使用：

row.extend(x[1] for x in emails if x[0] == row[1])

单线解决方案

或者，消除循环并将代码全部压缩成一行：

>>> projects = [  row + [x[1] for x in emails if x[0] == row[1]] for row in projects ]
>>> print projects 
[['project1', 'email1', 'john@gmail.com'], ['project1', 'email2', 'bill@gmail.com']]

Answer 2

emails = [['email1','john@gmail.com'],['email2','bill@gmail.com']]

projects = [['project1', 'email1'], ['project1', 'email2']]
from itertools import chain


print([list(set((chain.from_iterable(ele)))) for ele in zip(emails,projects)])
[['email1', 'john@gmail.com', 'project1'], ['email2', 'project1', 'bill@gmail.com']]

或者：

print([list(set(ele).union(projects[ind])) for ind, ele in enumerate(emails)])

或者：

print([projects[ind] + [ele for ele in sub if ele not in projects[ind]] for ind, sub in enumerate(emails)])

所有不同的版本都适用于多个项目，而不仅仅是针对单个元素进行检查。

如何组合两个列表

2 个答案:

单线解决方案