Question

SyntaxError：'yield'在函数外部

>>> for x in range(10):
...     yield x*x
... 
  File "<stdin>", line 2
SyntaxError: 'yield' outside function

我该怎么办？当我尝试在for循环中使用简单的yield时。

Answer 1

修改

您在评论中引用了scala，所以我认为您可能是在列表理解之后：

>>> squares = [i*i for i in range(10)] >>> squares [0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

您还可以使用生成器表达式：

>>> squares = (i*i for i in range(10)) >>> squares <generator object <genexpr> at 0x7f5299e04690> >>> list(squares) [0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

您需要在函数中调用yield。这使得该函数成为生成器函数。然后，您可以迭代函数产生的连续值，例如：

def squares(N): for i in range(N): yield i*i >>> squares(10) <generator object squares at 0x7f5299e04500> >>> for n in squares(10): ... print(n) 0 1 4 9 16 25 36 49 64 81 >>> list(squares(100)) [0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481, 3600, 3721, 3844, 3969, 4096, 4225, 4356, 4489, 4624, 4761, 4900, 5041, 5184, 5329, 5476, 5625, 5776, 5929, 6084, 6241, 6400, 6561, 6724, 6889, 7056, 7225, 7396, 7569, 7744, 7921, 8100, 8281, 8464, 8649, 8836, 9025, 9216, 9409, 9604, 9801]

Answer 2

注意：此语法在 Python 3.7 中已弃用，并将在 Python 3.8 中引发 SyntaxError

lamyield = lambda: [(yield x*x) for x in range(15)]
print(*lamyield())

另一种方式，

lanyield = lambda: (yield from (i ** 2 for i in range(15)))
for i in lanyield():
    print(i)

0
1
4
9
16
25
36
49
64
81
100
121
144
169
196

[Program finished]

如何在python中使用yield函数

2 个答案: