将ints格式化为十六进制的字符串 [英] Format ints into string of hex
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问题描述
我需要从随机整数列表(0-255)创建一个十六进制数字的字符串。每个十六进制数字都应该用两个字符表示:5 - 05,16 - 10等。
示例:
输入:[0,1,2,3,127,200,255],
输出:000102037fc8ff
$ c $
$ b
pre>#!/ usr / bin / env python
def format_me(nums):
result =
for i in nums :
如果i <= 9:
result + =0%x%i
else:
result + =%x%i
return结果
print format_me([0,1,2,3,127,200,255])
但是,这看起来有点尴尬。有没有更简单的方法?
解决方案
''。join('%02x'我为我输入)
I need to create a string of hex digits from a list of random integers (0-255). Each hex digit should be represented by two characters: 5 - "05", 16 - "10", etc.
Example:
Input: [0,1,2,3,127,200,255], Output: 000102037fc8ff
I've managed to come up with:
#!/usr/bin/env python def format_me(nums): result = "" for i in nums: if i <= 9: result += "0%x" % i else: result += "%x" % i return result print format_me([0,1,2,3,127,200,255])
However, this looks a bit awkward. Is there a simpler way?
解决方案''.join('%02x'%i for i in input)
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