Python one-liner to convert float to string

Question

I want to convert a float between 0.0 and 39.9 into a string. Replace the tens digit with an L, T or Y if it's a 1, 2 or 3 respectively. And append an M if it's in the ones. For example, 22.3 would return T2.3 and 8.1 would return M8.1 and so forth. Otherwise, return the float.

This code works of course, but I wondering if there was a simpler (if not one-liner) solution. Here's the code:

def specType(SpT):
  if 0 <= SpT <= 9.9:
    return 'M{}'.format(SpT)
  elif 10.0 <= SpT <= 19.9:
    return 'L{}'.format(SpT - 10)
  elif 20.0 <= SpT <= 29.9:
    return 'T{}'.format(SpT - 20)
  elif 30.0 <= SpT <= 39.9:
    return 'Y{}'.format(SpT - 30) 
  else:
    return SpT

Thanks!

You've got to be careful with all those float comparisons (SpT <= 9.9 etc) — NPE
– NPE, Commented Dec 17, 2012 at 19:49
+1, NPE. What is the desired output when SpT is 9.95? Because it's greater than 9.9 and less than 10.0, it falls through the cracks and reaches the return SpT line. — Kevin
– Kevin, Commented Dec 17, 2012 at 19:53
That's a good point. It's best if I account for that but in truth, it will never come up in the application. — Joe Flip
– Joe Flip, Commented Dec 18, 2012 at 15:42

DSM · Accepted Answer · 2012-12-17 19:58:23Z

10

How about:

def specType(SpT):
    return '{}{}'.format('MLTY'[int(SpT//10)], SpT % 10) if 0.0 <= SpT <= 39.9 else SpT

which gives

>>> specType(0.0)
'M0.0'
>>> specType(8.1)
'M8.1'
>>> specType(14.5)
'L4.5'
>>> specType(22.3)
'T2.3'
>>> specType(34.7)
'Y4.7'

[As noted in the comments, you'll want to think about what to do with numbers which can sneak through your boundaries -- I made one guess; modify as appropriate.]

edited Dec 17, 2012 at 19:58

answered Dec 17, 2012 at 19:51

DSM

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4 Comments

jackcogdill Over a year ago

:O Mind = Blown! I never knew you could do //!

Lev Levitsky Over a year ago

@yentup You can also use divmod :)

Lev Levitsky Over a year ago

@yentup That't a good read :) Another two from my bookmark bar: built-in types and data model (the latter may be a little more advanced).

Joe Flip Over a year ago

Fantastic! That's exactly it. Thanks!

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