Question

我有一个列表：for i in range(len(fruits)): for j range(len(fruits[i+1:])): print i,j

如何创建循环，使一个索引依赖于另一个索引：

'apple', 'orange'
'orange', 'blueberry'
'blueberry', strawberry'
'orange', 'blueberry'
etc...

我想打印出对：

 for(i=0;i<5;i++) 
     for (j=i+1; j<5; j++)
         print i, j

我想获得与c ++语言相对应的循环：

<ul>
  <li class="menu" data-banner-text="A SELECTED"><a>A</a></li>
  <li class="menu" data-banner-text="B SELECTED"><a>B</a></li>
  <li class="menu" data-banner-text="C SELECTED"><a>C</a></li>
  <li class="menu" data-banner-text="D SELECTED"><a>D</a></li>
</ul>
<div class="banner-header">
    TEST
</div>

Answer 1

如果您想要打印出C ++代码，请使用itertools.combinations：

In [1]: import itertools

In [3]: fruits = ['apple', 'orange', 'blueberry', 'strawberry']

In [4]: for res in itertools.combinations(fruits, 2):
   ...:     print res
   ...:
('apple', 'orange')
('apple', 'blueberry')
('apple', 'strawberry')
('orange', 'blueberry')
('orange', 'strawberry')
('blueberry', 'strawberry')

Answer 2

根据您的输出，我将使用此

fruits = ['apple', 'orange', 'blueberry', 'strawberry']
l = len(fruits)

for i in range(l):
   for j in range(i, l - 1):
       print fruits[j], fruits[j + 1]

输出：

apple orange
orange blueberry
blueberry strawberry
orange blueberry
blueberry strawberry
blueberry strawberry

Answer 3

for i in range(len(fruits)-1):
   print fruits[i], fruits[i+1]

Answer 4

如果我理解正确，请尝试：

>>> fruits = ['apple', 'orange', 'blueberry', 'strawberry']
>>> from itertools import combinations
>>> list(combinations(fruits,2))
[('apple', 'orange'), ('apple', 'blueberry'), ('apple', 'strawberry'), ('orange', 'blueberry'), ('orange', 'strawberry'), ('blueberry', 'strawberry')]

或者，只是Python化您的C循环：

>>> for i in range(0, len(fruits)):
...     for j in range(i+1, len(fruits)):
...        print fruits[i], fruits[j]
... 
apple orange
apple blueberry
apple strawberry
orange blueberry
orange strawberry
blueberry strawberry

循环索引取决于前一个索引

4 个答案: