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