a = np.array([0.123456789121212,2,3], dtype=np.float16)
print("16bit: ", a[0])
a = np.array([0.123456789121212,2,3], dtype=np.float32)
print("32bit: ", a[0])
b = np.array([0.123456789121212121212,2,3], dtype=np.float64)
print("64bit: ", b[0])
- 16bit: 0.1235
- 32bit: 0.12345679
- 64bit: 0.12345678912121212
python - The real difference between float32 and float64 - Stack Overflow
If float32 is only 6-7 digits, how do values where the first 6-7 digits are just 0s make sense?
floating point - What is the range of values a float can have in Python? - Stack Overflow
python - range() for floats - Stack Overflow
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a = np.array([0.123456789121212,2,3], dtype=np.float16)
print("16bit: ", a[0])
a = np.array([0.123456789121212,2,3], dtype=np.float32)
print("32bit: ", a[0])
b = np.array([0.123456789121212121212,2,3], dtype=np.float64)
print("64bit: ", b[0])
- 16bit: 0.1235
- 32bit: 0.12345679
- 64bit: 0.12345678912121212
float32 is a 32 bit number - float64 uses 64 bits.
That means that float64’s take up twice as much memory - and doing operations on them may be a lot slower in some machine architectures.
However, float64’s can represent numbers much more accurately than 32 bit floats.
They also allow much larger numbers to be stored.
For your Python-Numpy project I'm sure you know the input variables and their nature.
To make a decision we as programmers need to ask ourselves
- What kind of precision does my output need?
- Is speed not an issue at all?
- what precision is needed in parts per million?
A naive example would be if I store weather data of my city as [12.3, 14.5, 11.1, 9.9, 12.2, 8.2]
Next day Predicted Output could be of 11.5 or 11.5164374
do your think storing float 32 or float 64 would be necessary?
I was messing around with float32 numbers and got values like 1.4e-44. I'm not good at scientific notation, but I printed variables out with 100 digits being shown instead of scientific notation, and it goes something like .000000000000000000000000125321 (not exact, just an example)
But how does that make sense if float32 only allows 6-7 digits? Wouldn't that mean only the 0s are valid?
>>> import sys
>>> sys.float_info
sys.float_info(max=1.7976931348623157e+308, max_exp=1024, max_10_exp=308,
min=2.2250738585072014e-308, min_exp=-1021, min_10_exp=-307, dig=15,
mant_dig=53, epsilon=2.2204460492503131e-16, radix=2, rounds=1)
The smallest is sys.float_info.min (2.2250738585072014e-308) and the biggest is sys.float_info.max (1.7976931348623157e+308). See documentation for other properties.
sys.float_info.min is the normalized min. You can usually get the denormalized min as sys.float_info.min * sys.float_info.epsilon. Note that such numbers are represented with a loss of precision. As expected, the denormalized min is less than the normalized min.
See this post.
Relevant parts of the post:
In [2]: import kinds In [3]: kinds.default_float_kind.M kinds.default_float_kind.MAX kinds.default_float_kind.MIN kinds.default_float_kind.MAX_10_EXP kinds.default_float_kind.MIN_10_EXP kinds.default_float_kind.MAX_EXP kinds.default_float_kind.MIN_EXP In [3]: kinds.default_float_kind.MIN Out[3]: 2.2250738585072014e-308
You can either use:
[x / 10.0 for x in range(5, 50, 15)]
or use lambda / map:
map(lambda x: x/10.0, range(5, 50, 15))
def frange(x, y, jump):
while x < y:
yield x
x += jump
---
As the comments mention, this could produce unpredictable results like:
>>> list(frange(0, 100, 0.1))[-1]
99.9999999999986
To get the expected result, you can use one of the other answers in this question, or as @Tadhg mentioned, you can use decimal.Decimal as the jump argument. Make sure to initialize it with a string rather than a float.
>>> import decimal
>>> list(frange(0, 100, decimal.Decimal('0.1')))[-1]
Decimal('99.9')
Or even:
import decimal
def drange(x, y, jump):
while x < y:
yield float(x)
x += decimal.Decimal(jump)
And then:
>>> list(drange(0, 100, '0.1'))[-1]
99.9
[editor's not: if you only use positive jump and integer start and stop (x and y) , this works fine. For a more general solution see here.]
Quoting from a numpy discussion list:
That information is available via
numpy.finfo()andnumpy.iinfo():In [12]: finfo('d').max Out[12]: 1.7976931348623157e+308 In [13]: iinfo('i').max Out[13]: 2147483647 In [14]: iinfo('uint8').max Out[14]: 255
Link here.
You can use numpy.iinfo(arg).max to find the max value for integer types of arg, and numpy.finfo(arg).max to find the max value for float types of arg.
>>> numpy.iinfo(numpy.uint64).min
0
>>> numpy.iinfo(numpy.uint64).max
18446744073709551615L
>>> numpy.finfo(numpy.float64).max
1.7976931348623157e+308
>>> numpy.finfo(numpy.float64).min
-1.7976931348623157e+308
iinfo only offers min and max, but finfo also offers useful values such as eps (the smallest number > 0 representable) and resolution (the approximate decimal number resolution of the type of arg).