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up vote 4 down vote favorite

My background is in C, and I'm finally learning this new-fangled "Python" that all the cool kids are talking about.

I want to create a list of formatted hexadecimal values like so:

['0x00','0x01','0x02','0x03' (...) '0xfe','0xff']

Please note that the leading zero's shouldn't be stripped. For example, I want 0x03, not 0x3.

My first successful attempt was this:

hexlist=list()
tempbytes = bytes(range(256))
for value in tempbytes:
    hexlist.append("0x" + tempbytes[value:value+1].hex())
del tempbytes

But, wow this is ugly. Then I tried to make it more Pythonic ("Pythony"?) like so:

hexlist = ["0x"+bytes(range(256))[x:x+1].hex() for x in bytes(range(256))]

My thoughts:

  1. OMG! That's harder to read, not easier!
  2. Not only that, but I had to invoke range twice, which I assume is inefficient.
  3. I bet there's a much better way that I haven't been exposed to yet...

My questions:

  1. Is the second much less efficient?
  2. Is there a better way to do this?
  3. What's the best thing to do here in keeping with python style?
share|improve this question
up vote 3 down vote accepted

The shortest and most readable way to do it would probably be

hexlist = [hex(x) for x in range(256)]

Time

Your first : 0.1 ms per loop for \$10^5\$ loops

Your second: 1 ms per loop for \$10^5\$ loops

The above : 0.02 ms per loop for \$10^5\$ loops

I'm unable to com up with a worse example, but there is always a worse way to do it, exactly as there is always preferable way to do it.

Edit To do this with filled zeros I would suggest to treat the first 10 numbers as a special case.

hexlist = [hex(x) if x > 15 else "{:#04x}".format(x) for x in range(256)]

new time: 0.04 ms. Even if it isn't as pretty.

Edit:

And when considering that format can format binary numbers, why not hex?

hexlist = ["{:#04x}".format(x) for x in range(256)]

I guess that this is hard coded for this range, but anyway.

Time: 0.1ms for per loop for \$10^5\$ loops.

And more edits

Using old style:.

hexlist = ["0x%02x" % n for n in range(256)]

Time: 0.08ms per loop for \$10^5\$ loops.

And specifically generating what we want.

hexlist = ["{:#04x}".format(x) for x in range(16)] + \
              [hex(x) for x in range(16, 256)]

Time: 0.03ms per loop for \$10^5\$ loops.

share|improve this answer
    
Excellent, Simon, thank you. Unfortunately, this strips off the leading zeros from the first 10 values. I wasn't clear in my question: I would like 0x03, for example, instead of 0x3. I've updated the question... – bitsmack yesterday
    
Maybe not as pretty as your first method, but a whole lot prettier than either of mine! :-D Thanks! – bitsmack yesterday
    
Glad to help, it was an very interesting question! – Simon yesterday
    
For roughly 0.13 ms per loop (over 10^5 loops), using "old-style" formatting strings, ["0x%02x" % n for in in range(256)] is short, readable and (probably) fast enough for once-off generation. – Vatine yesterday
1  
Both your edits are using the Format Specification Mini-Language, format(x, "#04x") is the same as "{:#04x}".format(x). Also either way you could change them to '0x{:0>{}x}'.format(5, 2), or "{:#0{}x}".format(5, 4) if you don't want a static length. – Peilonrayz yesterday

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