在python中的新行上打印一些元素

我有一份清单：

a = ["748","994","530","47","178","908","374","188","88","78","56","93","30",
     "197","39","0","160","205","69","98","58","55","50","31","19","13","11",
     "20","20","15","650","175","259","87","193","59","48","47","19","18",
     "30","14","21"]

如何在一行上打印前17个元素，在新行上打印后17个元素，依此类推到最后？如果可以创建函数，而不是使用：

print a[:18]
print[18:36]
....
print [n-17:n+1]

之后我想把它们转换成数字

谢谢

接受可选的开始、停止、可选步骤参数：

使用它，您可以获得起始索引：0、17、34等等

这样，使用切片表示法，就可以得到所需的子列表

>>> a = ["748","994","530","47","178","908","374","188","88","78","56","93", ...]

>>> for i in range(0, len(a), 17):
...     print a[i:i+17]
...
['748', '994', '530', '47', '178', '908', '374', '188', '88', '78', '56', '93', '30', '197', '39', '0', '160']
['205', '69', '98', '58', '55', '50', '31', '19', '13', '11', '20', '20', '15', '650', '175', '259', '87']
['193', '59', '48', '47', '19', '18', '30', '14', '21']

>>> for i in range(0, len(a), 17):
...     print ' '.join(a[i:i+17])
...
748 994 530 47 178 908 374 188 88 78 56 93 30 197 39 0 160
205 69 98 58 55 50 31 19 13 11 20 20 15 650 175 259 87
193 59 48 47 19 18 30 14 21

这可能会有帮助

print "\n".join(str(a[i:i+17])[1:-1] for i in range(0,len(a),17))

---

看起来您已经开始学习python了。我强烈建议您研究for循环。还可以查看xrange/range，它们可以很容易地查找超过个数的数据。xrange和range使用格式范围[start、]end、[[skip]]。range和xrange在开始时都是包含的，但在结束时是独占的。括号用于表示可选输入。因此，仅提供1个输入实际上意味着范围0，结束，1。提供两个输入意味着rangestart、end、1。最后，提供3个输入意味着范围开始、结束和跳过。还要记住，xrange只在循环中使用，因为它实际上不生成列表，所以xrange被认为是迭代器

xrange/range的示例：

>>> range(10)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> range(5,10)
[5, 6, 7, 8, 9]
>>> range(5,10,2)
[5, 7, 9]

为了将变量转换为不同的类型，可以强制转换它们。因此，intn将n转换为整数。同样，strs将把s转换成字符串。试图将一个变量转换成它不能被转换的东西将抛出异常。例如：在下面的世界里！不会工作，因为你好世界！作为一个整数没有意义

# this function will print out n elements per line
# use printNElements(a, 17) to print out 17 at a time
def printNElements(a, n):
    for index in xrange(0,len(a),n):
        print a[index:index+n]

# this converts all the strings in list a to integers
a = [int(n) for n in a]