相关疑难解决方法(0)

如何访问索引本身以获取如下列表？

ints = [8, 23, 45, 12, 78]
for i in ints:
    print('item #{} = {}'.format(???, i))

当我使用循环遍历它时for,如何访问循环索引,在这种情况下从1到5？

我有一个2D列表,如下所示:

table = [['donkey', '2', '1', '0'], ['goat', '5', '3', '2']]

我想将最后三个元素更改为整数,但下面的代码感觉非常难看:

for row in table:
    for i in range(len(row)-1):
        row[i+1] = int(row[i+1])

但我宁愿有一些看起来像:

for row in table:
    for col in row[1:]:
        col = int(col)

我认为应该有一种方法来编写上面的代码,但是切片创建了一个与原始代码分开的迭代器/新列表,因此引用不会延续.

有没有办法获得更多的Pythonic解决方案？

如果我只想在循环中使用索引,我应该更好地range/xrange结合使用该函数len()

a = [1,2,3]
for i in xrange(len(a)):
    print i 

还是enumerate？即使我根本不使用p？

for i,p in enumerate(a):
    print i    

I read from range(len(list)) or enumerate(list)? that using range(len(s)) is not very good way to write Python. How one can write for loops in alternative way if we do not need to loop len(s) times but for example len(s)//3 times or len(s)-5 times? Is it possible to convert those loops to use enumerate?

For example, I had a project where I had a list of 3n elements 's[0], s[1],...,s[3n-1]' and I needed to print them in a nx3 table. …