This is really easy to do if you use the matplotlib.ticker.FormatStrFormatter as opposed to the LogFormatter. The following code will label everything with the format ‘%.2e’: import numpy as np import matplotlib.pyplot as plt import matplotlib.ticker as mtick fig = plt.figure() ax = fig.add_subplot(111) x = np.linspace(0, 300, 20) y = np.linspace(0,300, 20) y = … Read more
To round to d significant digits: >> d = 3; %// number of digits >> x = 5.237234; %// example number >> D = 10^(d-ceil(log10(x))); >> y = round(x*D)/D y = 5.2400 To round to d decimal digits: >> d = 3; %// number of digits >> x = 5.237234; %// example number >> D … Read more
These are talking about two slightly different things. The 7.2251 digits is the precision with which a number can be stored internally. For one example, if you did a computation with a double precision number (so you were starting with something like 15 digits of precision), then rounded it to a single precision number, the … Read more
This is really easy to do if you use the matplotlib.ticker.FormatStrFormatter as opposed to the LogFormatter. The following code will label everything with the format ‘%.2e’: import numpy as np import matplotlib.pyplot as plt import matplotlib.ticker as mtick fig = plt.figure() ax = fig.add_subplot(111) x = np.linspace(0, 300, 20) y = np.linspace(0,300, 20) y = … Read more
See: RoundToSignificantFigures by “P Daddy”. I’ve combined his method with another one I liked. Rounding to significant figures is a lot easier in TSQL where the rounding method is based on rounding position, not number of decimal places – which is the case with .Net math.round. You could round a number in TSQL to negative … Read more
Here’s the same code in Java without the 12.100000000000001 bug other answers have I also removed repeated code, changed power to a type integer to prevent floating issues when n – d is done, and made the long intermediate more clear The bug was caused by multiplying a large number with a small number. Instead … Read more
The framework doesn’t have a built-in function to round (or truncate, as in your example) to a number of significant digits. One way you can do this, though, is to scale your number so that your first significant digit is right after the decimal point, round (or truncate), then scale back. The following code should … Read more
You can use negative numbers to round integers: >>> round(1234, -3) 1000.0 Thus if you need only most significant digit: >>> from math import log10, floor >>> def round_to_1(x): … return round(x, -int(floor(log10(abs(x))))) … >>> round_to_1(0.0232) 0.02 >>> round_to_1(1234243) 1000000.0 >>> round_to_1(13) 10.0 >>> round_to_1(4) 4.0 >>> round_to_1(19) 20.0 You’ll probably have to take care … Read more