For non-negative (unsigned) integers only, use isdigit(): >>> a = “03523” >>> a.isdigit() True >>> b = “963spam” >>> b.isdigit() False Documentation for isdigit(): Python2, Python3 For Python 2 Unicode strings: isnumeric().
If the idea is to print integers stored as doubles as if they are integers, and otherwise print the doubles with the minimum necessary precision: public static String fmt(double d) { if(d == (long) d) return String.format(“%d”,(long)d); else return String.format(“%s”,d); } Produces: 232 0.18 1237875192 4.58 0 1.2345 And does not rely on string manipulation.
Summary In Java 6 (and presumably earlier), round(x) is implemented as floor(x+0.5).1 This is a specification bug, for precisely this one pathological case.2 Java 7 no longer mandates this broken implementation.3 The problem 0.5+0.49999999999999994 is exactly 1 in double precision: static void print(double d) { System.out.printf(“%016x\n”, Double.doubleToLongBits(d)); } public static void main(String args[]) { double … Read more
Use random.uniform(a, b): >>> import random >>> random.uniform(1.5, 1.9) 1.8733202628557872
Be extremely careful using any of the other suggestions. It all depends on context. I have spent a long time tracing bugs in a system that presumed a==b if |a-b|<epsilon. The underlying problems were: The implicit presumption in an algorithm that if a==b and b==c then a==c. Using the same epsilon for lines measured in … Read more
The math.ceil (ceiling) function returns the smallest integer higher or equal to x. For Python 3: import math print(math.ceil(4.2)) For Python 2: import math print(int(math.ceil(4.2)))
From the Floating-Point Guide: What can I do to avoid this problem? That depends on what kind of calculations you’re doing. If you really need your results to add up exactly, especially when you work with money: use a special decimal datatype. If you just don’t want to see all those extra decimal places: simply … Read more
In Python 2, division of two ints produces an int. In Python 3, it produces a float. We can get the new behaviour by importing from __future__. >>> from __future__ import division >>> a = 4 >>> b = 6 >>> c = a / b >>> c 0.66666666666666663
An explicit cast to double like this isn’t necessary: double trans = (double) trackBar1.Value / 5000.0; Identifying the constant as 5000.0 (or as 5000d) is sufficient: double trans = trackBar1.Value / 5000.0; double trans = trackBar1.Value / 5000d;
Rather than using a decimal step directly, it’s much safer to express this in terms of how many points you want. Otherwise, floating-point rounding error is likely to give you a wrong result. Use the linspace function from the NumPy library (which isn’t part of the standard library but is relatively easy to obtain). linspace … Read more