有人可以向我解释一下。
所以,我有一个列表s
,数字从-1到1,我想提取列表中某些数字的位置。
示例:
s= [-1, 0.5, 0.2, -0.9]
z = enumerate(s)
y1 = []
for i,j in z:
if j<=-0.8 and j>=-1:
k = i
y1.append(k)
y2 = []
for i,j in z:
if j<=0.8 and j>=-0.8:
k = i
y2.append(k)
我得到y1 = [0, 3]
和y2 = []
但如果我定义第二个enumerate
:
z1 = enumerate(s)
y1 = []
for i,j in z1:
if j<=-0.8 and j>=-1:
k = i
y1.append(k)
z2 = enumerate(s)
y2 = []
for i,j in z2:
if j<=0.8 and j>=-0.8:
k = i
y2.append(k)
我得到结果y1 = [0, 3]
和y2 = [1, 2]
为什么我需要第二个enumerate
?
答案 0 :(得分:3)
enumerate
返回序列的迭代器。遍历后,无法再次使用:
In [1]: l = [1, 2, 3, 4, 5]
In [2]: e = enumerate(l)
In [3]: e
Out[3]: <enumerate at 0x7fad7aea25e8>
In [4]: list(e)
Out[4]: [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]
In [5]: list(e)
Out[5]: []
你可以用一个列表包装它:
In [6]: e = list(enumerate(l))
In [7]: e
Out[7]: [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]
In [8]: list(e)
Out[8]: [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]
In [9]: list(e)
Out[9]: [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]
您还可以使用列表推导来简化您的解决方案:
y1 = [i for i, x in enumerate(s) if -1 <= x <= -0.8]
y2 = [i for i, x in enumerate(s) if -0.8 <= x <= 0.8]
答案 1 :(得分:2)
从python docs,枚举的实现如下:
def enumerate(collection):
'Generates an indexed series: (0,coll[0]), (1,coll[1]) ...'
i = 0
it = iter(collection)
while 1:
yield (i, it.next())
i += 1
因为它使用&#39; yield&#39;,它是一个生成器,当你在行中迭代它时存储状态
for i,j in z
要仅使用一个枚举,请将生成器的值转换为列表并重用列表:
s= [-1, 0.5, 0.2, -0.9]
z = list(enumerate(s))
y1 = []
for i,j in z:
if j<=-0.8 and j>=-1:
k = i
y1.append(k)
y2 = []
for i,j in z:
if j<=0.8 and j>=-0.8:
k = i
y2.append(k)
print(y1)
print(y2)
输出:
[0, 3]
[1, 2]
答案 2 :(得分:2)
enumerate
是一个懒惰的生成器,一旦被调用并一直遍历它就完成了。所以,我认为最好直接在for
循环中包含其调用:
s= [-1, 0.5, 0.2, -0.9]
y1 = []
for i,j in enumerate(s):
if -1 <= j <= -0.8:
y1.append(i)
y2 = []
for i,j in enumerate(s):
if -0.8 <= j <= 0.8:
y2.append(i)
旁注:
if j<=-0.8 and j>=-1:
可以替换为if -1 <= j <= -0.8: