[英]Enumerate function
誰可以給我解釋一下這個。
所以,我有一個列表s
用數字從-1到1,我想提取列表中的某些電話號碼的位置。
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 = []
但是如果我定義第二個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)
我得到結果y1 = [0, 3]
和y2 = [1, 2]
為什么我需要第二次enumerate
?
enumerate
返回序列的迭代器。 遍歷后,無法再次使用:
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]: []
你可以用一個列表包裝它:
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)]
您還可以使用列表推導來簡化您的解決方案:
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]
從python docs ,枚舉的實現如下:
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
因為它使用'yield',所以它是一個生成器,當你在行中迭代它時存儲狀態
for i,j in z
要僅使用一個枚舉,請將生成器的值轉換為列表並重用該列表:
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)
輸出:
[0, 3]
[1, 2]
enumerate
是一個惰性生成器,一旦被調用並遍歷它就完成了。 所以,我認為最好直接在for
循環中包含它的調用:
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)
附注:
if j<=-0.8 and j>=-1:
可以替換為if -1 <= j <= -0.8:
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