billiejoex wrote: > Hi all. I'd need to aproximate a given float number into the next (int) > bigger one. Because of my bad english I try to explain it with some example: > > 5.7 --> 6 > 52.987 --> 53 > 3.34 --> 4 > 2.1 --> 3 >
Have a look at math.ceil >>> import math >>> math.ceil(5.7) 6.0 Will McGugan -- http://www.willmcgugan.com "".join({'*':'@','^':'.'}.get(c,0) or chr(97+(ord(c)-84)%26) for c in "jvyy*jvyyzpthtna^pbz") -- http://mail.python.org/mailman/listinfo/python-list
