我如何访问指数,同时在一个序列上以一个为旋转的序列?
xs = [8, 23, 45]
for x in xs:
print("item #{} = {}".format(index, x))
所需的产量:
item #1 = 8
item #2 = 23
item #3 = 45
我如何访问指数,同时在一个序列上以一个为旋转的序列?
xs = [8, 23, 45]
for x in xs:
print("item #{} = {}".format(index, x))
所需的产量:
item #1 = 8
item #2 = 23
item #3 = 45
当前回答
for i in range(len(ints)):
print(i, ints[i]) # print updated to print() in Python 3.x+
其他回答
for i in range(len(ints)):
print(i, ints[i]) # print updated to print() in Python 3.x+
古老的风格:
for ix in range(len(ints)):
print(ints[ix])
列表理解:
[ (ix, ints[ix]) for ix in range(len(ints))]
>>> ints
[1, 2, 3, 4, 5]
>>> for ix in range(len(ints)): print ints[ix]
...
1
2
3
4
5
>>> [ (ix, ints[ix]) for ix in range(len(ints))]
[(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]
>>> lc = [ (ix, ints[ix]) for ix in range(len(ints))]
>>> for tup in lc:
... print(tup)
...
(0, 1)
(1, 2)
(2, 3)
(3, 4)
(4, 5)
>>>
你也可以尝试一下:
data = ['itemA.ABC', 'itemB.defg', 'itemC.drug', 'itemD.ashok']
x = []
for (i, item) in enumerate(data):
a = (i, str(item).split('.'))
x.append(a)
for index, value in x:
print(index, value)
产量是
0 ['itemA', 'ABC']
1 ['itemB', 'defg']
2 ['itemC', 'drug']
3 ['itemD', 'ashok']
您可以使用一个变量,例如计算,以计算列表中的元素数量:
ints = [8, 23, 45, 12, 78]
count = 0
for i in ints:
count = count + 1
print('item #{} = {}'.format(count, i))
一个简单的答案,使用一段时间的旋转:
arr = [8, 23, 45, 12, 78]
i = 0
while i < len(arr):
print("Item ", i + 1, " = ", arr[i])
i += 1
出口:
Item 1 = 8
Item 2 = 23
Item 3 = 45
Item 4 = 12
Item 5 = 78