Basically, I have a list of float numbers with too many decimals. So when I created a second list with two decimals, Python rounded them. I used the following:
g1= ["%.2f" % i for i in g]
Where g1 is the new list with two decimals, but rounded, and g is the list with float numbers.
How can I make one without rounding them?
I'm a newbie, btw. Thanks!
So, you want to truncate the numbers at the second digit?
Beware that rounding might be the better and more accurate solution anyway.
If you want to truncate the numbers, there are a couple of ways - one of them is to multiply the number by 10 elevated to the number of desired decimal places (100 for 2 places), apply "math.floor", and divide the total back by the same number.
However, as internal floating point arithmetic is not base 10, you'd risk getting more decimal places on the division to scale down.
Another way is to create a string with 3 digits after the "." and drop the last one - that'd be rounding proof.
And again, keep in mind that this converts the numbers to strings - what should be done for presentation purposes only. Also, "%" formatting is quite an old way to format parameters in a string. In modern Python, f-strings are the preferred way:
g1 = [f"{number:.03f}"[:-1] for number in g]
Another, more correct way, is, of course, treat numbers as numbers, and not play tricks on adding or removing digits on it. As noted in the comments, the method above would work for numbers like "1.227", that would be kept as "1.22", but not for "2.99999", which would be rounded to "3.000" and then truncated to "3.00".
Python has the decimal modules, which allows for arbitrary precision of decimal numbers - which includes less precision, if needed, and control of the way Python does the rounding - including rounding towards zero, instead of the nearest number.
Just set the decimal context to the decimal.ROUND_DOWN strategy, and then convert your numbers using either the round built-in (the exact number of digits is guaranteed, unlike using round with floating point numbers), or just do the rounding as part of the string formatting anyway. You can also convert your floats do Decimals in the same step:
from decimals import Decimal as D, getcontext, ROUND_DOWN
getcontext().rounding = ROUND_DOWN
g1 = [f"{D(number):.02f}" for number in g]
Again - by doing this, you could as well keep your numbers as Decimal objects, and still be able to perform math operations on them:
g2 = [round(D(number, 2)) for number in g]
Here is my solution where we don't even need to convert the number's to string to get the desired output:
def format_till_2_decimal(num):
return int(num*100)/100.0
g = [-5.427926, -12.222018, 7.214379, -16.771845, -6.1441464, 10.1383295, 14.740516, 5.9209185, -9.740783, -10.098338]
formatted_g = [format_till_2_decimal(num) for num in g]
print(formatted_g)
Hope this solution helps!!
Here might be the answer you are looking for:
g = [-5.427926, -12.222018, 7.214379, -16.771845, -6.1441464, 10.1383295, 14.740516, 5.9209185, -9.740783, -10.098338]
def trunc(number, ndigits=2):
parts = str(number).split('.') # divides number into 2 parts. for ex: -5, and 4427926
truncated_number = '.'.join([parts[0], parts[1][:ndigits]]) # We keep this first part, while taking only 2 digits from the second part. Then we concat it together to get '-5.44'
return round(float(truncated_number), 2) # This should return a float number, but to make sure it is roundded to 2 decimals.
g1 = [trunc(i) for i in g]
print(g1)
[-5.42, -12.22, 7.21, -16.77, -6.14, 10.13, 14.74, 5.92, -9.74, -10.09]
Hope this helps.
Actually if David's answer is what you are looking for, it can be done simply as following:
g = [-5.427926, -12.222018, 7.214379, -16.771845, -6.1441464, 10.1383295, 14.740516, 5.9209185, -9.740783, -10.098338]
g1 = [("%.3f" % i)[:-1] for i in g]
Just take 3 decimals, and remove the last chars from the result strings. (You may convert the result to float if you like)
Related
I have a random set of numbers in a SQL database:
1.2
0.4
5.1
0.0000000000232
1
7.54
0.000000000000006534
The decimals way below zero are displayed as scientific notation
num = 0.0000000000232
print(num)
> 2.23e-11
But that causes the rest of my code to bug out as the api behind it expects a decimal number. I checked it as I increased the precision with :.20f - that works fine.
Since the very small numbers are not constant with their precision, It would be unwise to simply set a static .20f.
What is a more elegant way to translate this to the correct decimal, always dynamic with the precision?
If Python provides a way to do this, they've hidden it very well. But a simple function can do it.
def float_to_str(x):
to_the_left = 1 + floor(log(x, 10))
to_the_right = sys.float_info.dig - to_the_left
if to_the_right <= 0:
s = str(int(x))
else:
s = format(x, f'0.{to_the_right}f').rstrip('0')
return s
>>> for num in [1.2, 0.4, 5.1, 0.0000000000232, 1, 7.54, 0.000000000000006534]:
print(float_to_str(num))
1.2
0.4
5.1
0.0000000000232
1.
7.54
0.000000000000006534
The first part uses the logarithm base 10 to figure out how many digits will be on the left of the decimal point, or the number of zeros to the right of it if the number is negative. To find out how many digits can be to the right, we take the total number of significant digits that a float can hold as given by sys.float_info.dig which should be 15 on most Python implementations, and subtract the digits on the left. If this number is negative there won't be anything but garbage after the decimal point, so we can rely on integer conversion instead - it never uses scientific notation. Otherwise we simply conjure up the proper string to use with format. For the final step we strip off the redundant trailing zeros.
Using integers for large numbers isn't perfect because we lose the rounding that naturally occurs with floating point string conversion. float_to_str(1e25) for example will return '10000000000000000905969664'. Since your examples didn't contain any such large numbers I didn't worry about it, but it could be fixed with a little more work. For the reasons behind this see Is floating point math broken?
e = str(2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274)
print(e)
Output:
2.718281828459045
Screenshots: here and here.
Why does the code only print out the first few characters of e instead of the whole string?
A string str has characters, but a number (be it an int or a float) just has a value.
If you do this:
e_first_100 = '2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274'
print(e_first_100)
You'll see all digits printed, because they are just characters in a string, it could have also been the first 100 characters from 'War and Peace' and you would not expect any of that to get lost either.
Since 'e' is not an integer value, you can't use int here, so you'll have to use float, but Python uses a finite number of bits to represent such a number, while there's an infinite number of real numbers. In fact there's an infinite number of values between any two real numbers. So a clever way has to be used to represent at least the ones you use most often, with a limited amount of precision.
You often don't notice the lack of precision, but try something like .1 + .1 + .1 == .3 in Python and you'll see that it can pop up in common situations.
Your computer already has a built-in way to represent these floating point numbers, using either 32 or 64 bits, although many languages (Python included) do offer additional ways of representing floats that aren't part of the way your computer works and allow a bit more precision. By default, Python uses these standard representations of real numbers.
So, if you then do this:
e1 = float(e_first_100)
print(e1)
e2 = 2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274
print(e2)
Both result in a value that, when you print it, looks like:
2.718281828459045
Because that's the precision up to which the number is (more or less) accurately represented.
If you need to use e in a more precise manner, you can use Python's own representation:
from decimal import Decimal
e3 = Decimal(e_first_100)
print(e3)
That looks promising, but even Decimal only has limited precision, although it's better than standard floats:
print(e2 * 3)
print(e3 * Decimal(3))
The difference:
8.154845485377136
8.154845485377135706080862414
To expand on Grismar's answer, you don't see the data because the default string representation of floats cuts off at that point as going further than that wouldn't be very useful, but while the object is a float the data is still there.
To get a string with the data, you could provide a fixed precision to some larger amount of digits, for example
In [2]: e = format(
...: 2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274,
...: ".50f",
...: )
In [3]: e
Out[3]: '2.71828182845904509079559829842764884233474731445312'
which gives us the first 50 digits, but this is of course not particularly useful with floats as the loss of precision picks up the further you go
I want to print floating point numbers with a set number of significant (i.e. non-zero) digits, but avoid the scientific notation (e).
So, for example, the number 0.000000002343245345 should be printed as 0.000000002343 (and not 2.343e-09)
I know how to define number of significant digits with an e notation:
>>>print('{:.3e}'.format(0.000000002343245345))
2.343e-09
And how to print a set number of decimal places without e-notation:
>>>print('{:.12f}'.format(0.000000002343245345))
0.000000002343
but not how to combine the two.
is there any simple way of doing so?
Here is some code that usually does what you want.
x = 0.000000002343245345
n = 4
from math import log10, floor
print('{:.{}f}'.format(x, n - floor(log10(x)) - 1))
Note that, due to the lack of exactness in floating-point arithmetic, that this may occasionally give a wrong answer. For example, the floor(log10()) may be one off from what is expected at or very near negative powers of 10 such as 0.1, 0.01, 0.001, and so on. My code seems to work well with those values but that is not guaranteed.
Also, there is no reasonable answer for some combinations of x and n. For example, what do you expect to result from x = 200000 and n = 4? There is no good answer, and my code gives the error
ValueError: Format specifier missing precision
You have to calculate the number of digits yourself. For four significant digits, this would be
number = 0.000000002343245345
digits = 4 - int(math.ceil(math.log10(number)))
print("{:.{}f}".format(number, digits))
# 0.000000002343
I would like to generate uniformly distributed random numbers between 0 and 0.5, but truncated to 2 decimal places.
without the truncation, I know this is done by
import numpy as np
rs = np.random.RandomState(123456)
set = rs.uniform(size=(50,1))*0.5
could anyone help me with suggestions on how to generate random numbers up to 2 d.p. only? Thanks!
A float cannot be truncated (or rounded) to 2 decimal digits, because there are many values with 2 decimal digits that just cannot be represented exactly as an IEEE double.
If you really want what you say you want, you need to use a type with exact precision, like Decimal.
Of course there are downsides to doing that—the most obvious one for numpy users being that you will have to use dtype=object, with all of the compactness and performance implications.
But it's the only way to actually do what you asked for.
Most likely, what you actually want to do is either Joran Beasley's answer (leave them untruncated, and just round at print-out time) or something similar to Lauritz V. Thaulow's answer (get the closest approximation you can, then use explicit epsilon checks everywhere).
Alternatively, you can do implicitly fixed-point arithmetic, as David Heffernan suggests in a comment: Generate random integers between 0 and 50, keep them as integers within numpy, and just format them as fixed point decimals and/or convert to Decimal when necessary (e.g., for printing results). This gives you all of the advantages of Decimal without the costs… although it does open an obvious window to create new bugs by forgetting to shift 2 places somewhere.
decimals are not truncated to 2 decimal places ever ... however their string representation maybe
import numpy as np
rs = np.random.RandomState(123456)
set = rs.uniform(size=(50,1))*0.5
print ["%0.2d"%val for val in set]
How about this?
np.random.randint(0, 50, size=(50,1)).astype("float") / 100
That is, create random integers between 0 and 50, and divide by 100.
EDIT:
As made clear in the comments, this will not give you exact two-digit decimals to work with, due to the nature of float representations in memory. It may look like you have the exact float 0.1 in your array, but it definitely isn't exactly 0.1. But it is very very close, and you can get it closer by using a "double" datatype instead.
You can postpone this problem by just keeping the numbers as integers, and remember that they're to be divided by 100 when you use them.
hundreds = random.randint(0, 50, size=(50, 1))
Then at least the roundoff won't happen until at the last minute (or maybe not at all, if the numerator of the equation is a multiple of the denominator).
I managed to find another alternative:
import numpy as np
rs = np.random.RandomState(123456)
set = rs.uniform(size=(50,2))
for i in range(50):
for j in range(2):
set[i,j] = round(set[i,j],2)
I have two sets of data that I am reading via nested for loops in Python. I need to match lines of the two different text files using a common number (time). In the two files, time is written differently (ex. 21:53:28.339 vs. 121082008.3399). I only need the last four digits of the times to match them up, for example from 21:53:28.339, I only need '8.339'. For the most part, indexing the number as a string works just fine (ex: timeList[nid][7:]), except for situations such as the numbers listed above, where python rounds .3399 to .34.
Is there a way for me to keep the numbers in float form and to select unrounded digits from the data?
Thanks!
edit - using Decimal exclusively - with full example
import decimal
def simplify(text):
# might be a : separated value
text = text.split(':')[-1]
# turn into decimal
number = decimal.Decimal(text)
# remove everything but the ones place and forwards
number = number - (number/10).quantize(1, rounding=decimal.ROUND_FLOOR) * 10
# truncate to the thousandths
return number.quantize(decimal.Decimal('.001'), rounding=decimal.ROUND_FLOOR)
a = '121082008.3399'
b = '21:53:28.339'
assert simplify(a) == simplify(b)
print simplify(a), '=', simplify(b)
Scott if you compare the numbers using strings then you don't need any floats and there will be no 'rounding' going on.
'8.339' == '8.339'
or, if you have
a = '8.3399'
b = '8.339'
then
a[:-1] == b
however if you do decide to work with them as 'numbers', then as Ignacio pointed out, you can use decimals.
from decimal import Decimal
number_a = Decimal(a[:-1])
number_b = Decimal(b)
now
number_a == number_b
Hope that helps
It appears from your description that you want to compare using one digit before the decimal point and 3 digits after the decimal point, using truncation instead of rounding. So just do that:
>>> def extract(s):
... i = s.find('.')
... return s[i-1:i+4]
...
>>> map(extract, ['21:53:28.339', '121082008.3399'])
['8.339', '8.339']
>>>
Use decimal.Decimal instead of float.