Format it yourself (see Format Specification Mini-Language): for ix in [.02e9, .2e9, 2e9, 20e9, 200e9, 2000e9]: print(‘{:.3e} => {:0=8.3f}e9’.format(ix, ix / 1e9)) Output 2.000e+07 => 0000.020e9 2.000e+08 => 0000.200e9 2.000e+09 => 0002.000e9 2.000e+10 => 0020.000e9 2.000e+11 => 0200.000e9 2.000e+12 => 2000.000e9 Explanation {:0=8.3f} means “zero-pad, pad between the sign and the number, total field width … Read more
This is what you need: np.set_printoptions(suppress=True) Here is the documentation.
You can do something like this: a = 200 a.toExponential(); //output 2e+2 fiddle: http://jsfiddle.net/Q8avJ/9/
You could use colorbar‘s format parameter: import matplotlib.pyplot as plt import numpy as np import matplotlib.ticker as ticker img = np.random.randn(300,300) myplot = plt.imshow(img) def fmt(x, pos): a, b = ‘{:.2e}’.format(x).split(‘e’) b = int(b) return r’${} \times 10^{{{}}}$’.format(a, b) plt.colorbar(myplot, format=ticker.FuncFormatter(fmt)) plt.show()
Double.Parse(“1.234567E-06”, System.Globalization.NumberStyles.Float);
Use std::fixed stream manipulator: cout<<fixed<<“Bas ana: “<<x<<“\tSon faiz: “<<t<<“\tSon ana: “<<x+t<<endl;
To set formatting of floating variables you can use a combination of setprecision(n), showpoint and fixed. In order to use parameterized stream manipulators like setprecision(n) you will have to include the iomanip library: #include <iomanip> setprecision(n): will constrain the floating-output to n places, and once you set it, it is set until you explicitly unset … Read more
You can effectively remove scientific notation in printing with this code: options(scipen=999)
System.out.println(String.format(“%6.3e”,223.45654543434)); results in 2.235e+02 which is the closest I get. more info : http://java.sun.com/j2se/1.5.0/docs/api/java/util/Formatter.html#syntax