How to write string (float) to file?

Jeff Shannon jeff at ccvcorp.com
Tue Aug 28 20:07:46 CEST 2001


Rainer Deyke wrote:

> <gbreed at cix.compulink.co.uk> wrote in message
>
> > c = '%.2f'%(float(a)/b)
>
> That's unreliable.  Use this instead:
>
> c = (a * 100) / b
> s = '%03d' % c
> file.write(s[:-2] + '.' + s[-2:])
>

Unreliable how?

>>> l1 = [5, 13, 23, 30, 42, 86, 101, 153, 999, 123456]
>>> l2 = [2, 5, 23, 59]
>>> def foo(a, b):
...  return '%.2f' % (float(a)/b)
...
>>> def bar(a, b):
...  s = '%03d' % ( (a*100)/b )
...  return s[:-2] + '.' + s[-2:]
...
>>> for b in l2:
...  for a in l1:
...   x, y = foo(a,b), bar(a,b)
...   full = repr(float(a)/b)
...   print '%10s %10s    %d  %s' % (x, y, (x==y), full)
...
      2.50       2.50    1  2.5
      6.50       6.50    1  6.5
     11.50      11.50    1  11.5
     15.00      15.00    1  15.0
     21.00      21.00    1  21.0
     43.00      43.00    1  43.0
     50.50      50.50    1  50.5
     76.50      76.50    1  76.5
    499.50     499.50    1  499.5
  61728.00   61728.00    1  61728.0
      1.00       1.00    1  1.0
      2.60       2.60    1  2.6000000000000001
      4.60       4.60    1  4.5999999999999996
      6.00       6.00    1  6.0
      8.40       8.40    1  8.4000000000000004
     17.20      17.20    1  17.199999999999999
     20.20      20.20    1  20.199999999999999
     30.60      30.60    1  30.600000000000001
    199.80     199.80    1  199.80000000000001
  24691.20   24691.20    1  24691.200000000001
      0.22       0.21    0  0.21739130434782608
      0.57       0.56    0  0.56521739130434778
      1.00       1.00    1  1.0
      1.30       1.30    1  1.3043478260869565
      1.83       1.82    0  1.826086956521739
      3.74       3.73    0  3.7391304347826089
      4.39       4.39    1  4.3913043478260869
      6.65       6.65    1  6.6521739130434785
     43.43      43.43    1  43.434782608695649
   5367.65    5367.65    1  5367.652173913043
      0.08       0.08    1  0.084745762711864403
      0.22       0.22    1  0.22033898305084745
      0.39       0.38    0  0.38983050847457629
      0.51       0.50    0  0.50847457627118642
      0.71       0.71    1  0.71186440677966101
      1.46       1.45    0  1.4576271186440677
      1.71       1.71    1  1.7118644067796611
      2.59       2.59    1  2.593220338983051
     16.93      16.93    1  16.932203389830509
   2092.47    2092.47    1  2092.4745762711864
>>>

The bar() version, your suggestion, appears to *usually* truncate, but in a few
cases rounds up.  The foo() version *always* rounds to nearest .01.  I would
consider the latter to be the more reliable (consistent) behavior.  (I was
surprised by this, actually--I expected bar() to always truncate, and was going
to simply point out that it was a matter of whether truncation or rounding was
preferred...)

Jeff Shannon
Technician/Programmer
Credit International





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