打印带有十六进制元素的 Python 列表

Question

I have this list:

a_list = [0x00, 0x00, 0x00, 0x00]

When I print it, I get:

print a_list
[0, 0, 0, 0]

But I want: [0x0, 0x0, 0x0, 0x0] or [0x00, 0x00, 0x00, 0x00] , it doesn't matter for now.

I've tried to create a function such as:

def hex_print(the_list):
     string = '['
     for element in the_list:
         if(the_list.index(element) < len(the_list)):
             print(str(the_list.index(element)))
             string = string + hex(element) + ', '
         else:
             print(str(the_list.index(element)))
             string = string + hex(element) + ']'
     print string

But every time the printed message is:

[0x0, 0x0, 0x0, 0x0,

I think that the_list.index(element) always returns the first occurrence of element in the_list and not the actual position of the element. Is there a way where I can get the actual position of the element?

Answer 1

>>> a_list = range(4)
>>> print '[{}]'.format(', '.join(hex(x) for x in a_list))
[0x0, 0x1, 0x2, 0x3]

This has the advantage of not putting quotes around all the elements, so it produces a valid literal for a list of numbers.

Answer 2

Try the following:

print [hex(x) for x in a_list]

The output shall be something like: http://codepad.org/56Vtgofl

Answer 3

print [hex(no) for no in a_list]

hex函数将数字转换为其十六进制表示。

Answer 4

对于python3，使用以下代码：

print([hex(x) for x in a_list])

Answer 5

If you want zero padding:

", ".join("0x{:04x}".format(num) for num in a_list)

The "0x{:04x}" puts the "0x" in front of the number. The "{:04x}" part pads zeros to 4 places, printing the number in hex.