I want to remove digits from a float to have a fixed number of digits after the dot, like:
1.923328437452 → 1.923
I need to output as a string to another function, not print.
Also I want to ignore the lost digits, not round them.
round(1.923328437452, 3)
See Python's documentation on the standard types. You'll need to scroll down a bit to get to the round function. Essentially the second number says how many decimal places to round it to.
First, the function, for those who just want some copy-and-paste code:
def truncate(f, n):
'''Truncates/pads a float f to n decimal places without rounding'''
s = '{}'.format(f)
if 'e' in s or 'E' in s:
return '{0:.{1}f}'.format(f, n)
i, p, d = s.partition('.')
return '.'.join([i, (d+'0'*n)[:n]])
This is valid in Python 2.7 and 3.1+. For older versions, it's not possible to get the same "intelligent rounding" effect (at least, not without a lot of complicated code), but rounding to 12 decimal places before truncation will work much of the time:
def truncate(f, n):
'''Truncates/pads a float f to n decimal places without rounding'''
s = '%.12f' % f
i, p, d = s.partition('.')
return '.'.join([i, (d+'0'*n)[:n]])
Explanation
The core of the underlying method is to convert the value to a string at full precision and then just chop off everything beyond the desired number of characters. The latter step is easy; it can be done either with string manipulation
i, p, d = s.partition('.')
'.'.join([i, (d+'0'*n)[:n]])
or the decimal module
str(Decimal(s).quantize(Decimal((0, (1,), -n)), rounding=ROUND_DOWN))
The first step, converting to a string, is quite difficult because there are some pairs of floating point literals (i.e. what you write in the source code) which both produce the same binary representation and yet should be truncated differently. For example, consider 0.3 and 0.29999999999999998. If you write 0.3 in a Python program, the compiler encodes it using the IEEE floating-point format into the sequence of bits (assuming a 64-bit float)
0011111111010011001100110011001100110011001100110011001100110011
This is the closest value to 0.3 that can accurately be represented as an IEEE float. But if you write 0.29999999999999998 in a Python program, the compiler translates it into exactly the same value. In one case, you meant it to be truncated (to one digit) as 0.3, whereas in the other case you meant it to be truncated as 0.2, but Python can only give one answer. This is a fundamental limitation of Python, or indeed any programming language without lazy evaluation. The truncation function only has access to the binary value stored in the computer's memory, not the string you actually typed into the source code.1
If you decode the sequence of bits back into a decimal number, again using the IEEE 64-bit floating-point format, you get
0.2999999999999999888977697537484345957637...
so a naive implementation would come up with 0.2 even though that's probably not what you want. For more on floating-point representation error, see the Python tutorial.
It's very rare to be working with a floating-point value that is so close to a round number and yet is intentionally not equal to that round number. So when truncating, it probably makes sense to choose the "nicest" decimal representation out of all that could correspond to the value in memory. Python 2.7 and up (but not 3.0) includes a sophisticated algorithm to do just that, which we can access through the default string formatting operation.
'{}'.format(f)
The only caveat is that this acts like a g format specification, in the sense that it uses exponential notation (1.23e+4) if the number is large or small enough. So the method has to catch this case and handle it differently. There are a few cases where using an f format specification instead causes a problem, such as trying to truncate 3e-10 to 28 digits of precision (it produces 0.0000000002999999999999999980), and I'm not yet sure how best to handle those.
If you actually are working with floats that are very close to round numbers but intentionally not equal to them (like 0.29999999999999998 or 99.959999999999994), this will produce some false positives, i.e. it'll round numbers that you didn't want rounded. In that case the solution is to specify a fixed precision.
'{0:.{1}f}'.format(f, sys.float_info.dig + n + 2)
The number of digits of precision to use here doesn't really matter, it only needs to be large enough to ensure that any rounding performed in the string conversion doesn't "bump up" the value to its nice decimal representation. I think sys.float_info.dig + n + 2 may be enough in all cases, but if not that 2 might have to be increased, and it doesn't hurt to do so.
In earlier versions of Python (up to 2.6, or 3.0), the floating point number formatting was a lot more crude, and would regularly produce things like
>>> 1.1
1.1000000000000001
If this is your situation, if you do want to use "nice" decimal representations for truncation, all you can do (as far as I know) is pick some number of digits, less than the full precision representable by a float, and round the number to that many digits before truncating it. A typical choice is 12,
'%.12f' % f
but you can adjust this to suit the numbers you're using.
1Well... I lied. Technically, you can instruct Python to re-parse its own source code and extract the part corresponding to the first argument you pass to the truncation function. If that argument is a floating-point literal, you can just cut it off a certain number of places after the decimal point and return that. However this strategy doesn't work if the argument is a variable, which makes it fairly useless. The following is presented for entertainment value only:
def trunc_introspect(f, n):
'''Truncates/pads the float f to n decimal places by looking at the caller's source code'''
current_frame = None
caller_frame = None
s = inspect.stack()
try:
current_frame = s[0]
caller_frame = s[1]
gen = tokenize.tokenize(io.BytesIO(caller_frame[4][caller_frame[5]].encode('utf-8')).readline)
for token_type, token_string, _, _, _ in gen:
if token_type == tokenize.NAME and token_string == current_frame[3]:
next(gen) # left parenthesis
token_type, token_string, _, _, _ = next(gen) # float literal
if token_type == tokenize.NUMBER:
try:
cut_point = token_string.index('.') + n + 1
except ValueError: # no decimal in string
return token_string + '.' + '0' * n
else:
if len(token_string) < cut_point:
token_string += '0' * (cut_point - len(token_string))
return token_string[:cut_point]
else:
raise ValueError('Unable to find floating-point literal (this probably means you called {} with a variable)'.format(current_frame[3]))
break
finally:
del s, current_frame, caller_frame
Generalizing this to handle the case where you pass in a variable seems like a lost cause, since you'd have to trace backwards through the program's execution until you find the floating-point literal which gave the variable its value. If there even is one. Most variables will be initialized from user input or mathematical expressions, in which case the binary representation is all there is.
The result of round is a float, so watch out (example is from Python 2.6):
>>> round(1.923328437452, 3)
1.923
>>> round(1.23456, 3)
1.2350000000000001
You will be better off when using a formatted string:
>>> "%.3f" % 1.923328437452
'1.923'
>>> "%.3f" % 1.23456
'1.235'
n = 1.923328437452
str(n)[:4]
At my Python 2.7 prompt:
>>> int(1.923328437452 * 1000)/1000.0
1.923
The truely pythonic way of doing it is
from decimal import *
with localcontext() as ctx:
ctx.rounding = ROUND_DOWN
print Decimal('1.923328437452').quantize(Decimal('0.001'))
or shorter:
from decimal import Decimal as D, ROUND_DOWN
D('1.923328437452').quantize(D('0.001'), rounding=ROUND_DOWN)
Update
Usually the problem is not in truncating floats itself, but in the improper usage of float numbers before rounding.
For example: int(0.7*3*100)/100 == 2.09.
If you are forced to use floats (say, you're accelerating your code with numba), it's better to use cents as "internal representation" of prices: (70*3 == 210) and multiply/divide the inputs/outputs.
Simple python script -
n = 1.923328437452
n = float(int(n * 1000))
n /=1000
def trunc(num, digits):
sp = str(num).split('.')
return '.'.join([sp[0], sp[1][:digits]])
This should work. It should give you the truncation you are looking for.
So many of the answers given for this question are just completely wrong. They either round up floats (rather than truncate) or do not work for all cases.
This is the top Google result when I search for 'Python truncate float', a concept which is really straightforward, and which deserves better answers. I agree with Hatchkins that using the decimal module is the pythonic way of doing this, so I give here a function which I think answers the question correctly, and which works as expected for all cases.
As a side-note, fractional values, in general, cannot be represented exactly by binary floating point variables (see here for a discussion of this), which is why my function returns a string.
from decimal import Decimal, localcontext, ROUND_DOWN
def truncate(number, places):
if not isinstance(places, int):
raise ValueError("Decimal places must be an integer.")
if places < 1:
raise ValueError("Decimal places must be at least 1.")
# If you want to truncate to 0 decimal places, just do int(number).
with localcontext() as context:
context.rounding = ROUND_DOWN
exponent = Decimal(str(10 ** - places))
return Decimal(str(number)).quantize(exponent).to_eng_string()
>>> from math import floor
>>> floor((1.23658945) * 10**4) / 10**4
1.2365
# divide and multiply by 10**number of desired digits
If you fancy some mathemagic, this works for +ve numbers:
>>> v = 1.923328437452
>>> v - v % 1e-3
1.923
I did something like this:
from math import trunc
def truncate(number, decimals=0):
if decimals < 0:
raise ValueError('truncate received an invalid value of decimals ({})'.format(decimals))
elif decimals == 0:
return trunc(number)
else:
factor = float(10**decimals)
return trunc(number*factor)/factor
You can do:
def truncate(f, n):
return math.floor(f * 10 ** n) / 10 ** n
testing:
>>> f=1.923328437452
>>> [truncate(f, n) for n in range(5)]
[1.0, 1.9, 1.92, 1.923, 1.9233]
Just wanted to mention that the old "make round() with floor()" trick of
round(f) = floor(f+0.5)
can be turned around to make floor() from round()
floor(f) = round(f-0.5)
Although both these rules break around negative numbers, so using it is less than ideal:
def trunc(f, n):
if f > 0:
return "%.*f" % (n, (f - 0.5*10**-n))
elif f == 0:
return "%.*f" % (n, f)
elif f < 0:
return "%.*f" % (n, (f + 0.5*10**-n))
def precision(value, precision):
"""
param: value: takes a float
param: precision: int, number of decimal places
returns a float
"""
x = 10.0**precision
num = int(value * x)/ x
return num
precision(1.923328437452, 3)
1.923
Short and easy variant
def truncate_float(value, digits_after_point=2):
pow_10 = 10 ** digits_after_point
return (float(int(value * pow_10))) / pow_10
>>> truncate_float(1.14333, 2)
>>> 1.14
>>> truncate_float(1.14777, 2)
>>> 1.14
>>> truncate_float(1.14777, 4)
>>> 1.1477
When using a pandas df this worked for me
import math
def truncate(number, digits) -> float:
stepper = 10.0 ** digits
return math.trunc(stepper * number) / stepper
df['trunc'] = df['float_val'].apply(lambda x: truncate(x,1))
df['trunc']=df['trunc'].map('{:.1f}'.format)
int(16.5);
this will give an integer value of 16, i.e. trunc, won't be able to specify decimals, but guess you can do that by
import math;
def trunc(invalue, digits):
return int(invalue*math.pow(10,digits))/math.pow(10,digits);
Here is an easy way:
def truncate(num, res=3):
return (floor(num*pow(10, res)+0.5))/pow(10, res)
for num = 1.923328437452, this outputs 1.923
def trunc(f,n):
return ('%.16f' % f)[:(n-16)]
A general and simple function to use:
def truncate_float(number, length):
"""Truncate float numbers, up to the number specified
in length that must be an integer"""
number = number * pow(10, length)
number = int(number)
number = float(number)
number /= pow(10, length)
return number
There is an easy workaround in python 3. Where to cut I defined with an help variable decPlace to make it easy to adapt.
f = 1.12345
decPlace= 4
f_cut = int(f * 10**decPlace) /10**decPlace
Output:
f = 1.1234
Hope it helps.
Most answers are way too complicated in my opinion, how about this?
digits = 2 # Specify how many digits you want
fnum = '122.485221'
truncated_float = float(fnum[:fnum.find('.') + digits + 1])
>>> 122.48
Simply scanning for the index of '.' and truncate as desired (no rounding).
Convert string to float as final step.
Or in your case if you get a float as input and want a string as output:
fnum = str(122.485221) # convert float to string first
truncated_float = fnum[:fnum.find('.') + digits + 1] # string output
I think a better version would be just to find the index of decimal point . and then to take the string slice accordingly:
def truncate(number, n_digits:int=1)->float:
'''
:param number: real number ℝ
:param n_digits: Maximum number of digits after the decimal point after truncation
:return: truncated floating point number with at least one digit after decimal point
'''
decimalIndex = str(number).find('.')
if decimalIndex == -1:
return float(number)
else:
return float(str(number)[:decimalIndex+n_digits+1])
int(1.923328437452 * 1000) / 1000
>>> 1.923
int(1.9239 * 1000) / 1000
>>> 1.923
By multiplying the number by 1000 (10 ^ 3 for 3 digits) we shift the decimal point 3 places to the right and get 1923.3284374520001. When we convert that to an int the fractional part 3284374520001 will be discarded. Then we undo the shifting of the decimal point again by dividing by 1000 which returns 1.923.
use numpy.round
import numpy as np
precision = 3
floats = [1.123123123, 2.321321321321]
new_float = np.round(floats, precision)
Something simple enough to fit in a list-comprehension, with no libraries or other external dependencies. For Python >=3.6, it's very simple to write with f-strings.
The idea is to let the string-conversion do the rounding to one more place than you need and then chop off the last digit.
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nout+1}f}'[:-1] for x in [2/3, 4/5, 8/9, 9/8, 5/4, 3/2]]
['0.666', '0.800', '0.888', '1.125', '1.250', '1.500']
Of course, there is rounding happening here (namely for the fourth digit), but rounding at some point is unvoidable. In case the transition between truncation and rounding is relevant, here's a slightly better example:
>>> nacc = 6 # desired accuracy (maximum 15!)
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nacc}f}'[:-(nacc-nout)] for x in [2.9999, 2.99999, 2.999999, 2.9999999]]
>>> ['2.999', '2.999', '2.999', '3.000']
Bonus: removing zeros on the right
>>> nout = 3 # desired number of digits in output
>>> [f'{x:.{nout+1}f}'[:-1].rstrip('0') for x in [2/3, 4/5, 8/9, 9/8, 5/4, 3/2]]
['0.666', '0.8', '0.888', '1.125', '1.25', '1.5']
The core idea given here seems to me to be the best approach for this problem.
Unfortunately, it has received less votes while the later answer that has more votes is not complete (as observed in the comments). Hopefully, the implementation below provides a short and complete solution for truncation.
def trunc(num, digits):
l = str(float(num)).split('.')
digits = min(len(l[1]), digits)
return l[0] + '.' + l[1][:digits]
which should take care of all corner cases found here and here.
Am also a python newbie and after making use of some bits and pieces here, I offer my two cents
print str(int(time.time()))+str(datetime.now().microsecond)[:3]
str(int(time.time())) will take the time epoch as int and convert it to string and join with...
str(datetime.now().microsecond)[:3] which returns the microseconds only, convert to string and truncate to first 3 chars
# value value to be truncated
# n number of values after decimal
value = 0.999782
n = 3
float(int(value*1en))*1e-n
I am writing a program that converts a decimal number into a binary. I made it that far and I stuck with a problem. In the code with print(aa).
I tried to get a binary form of given number but it prints 1. I think I have "return" function problem here how can I solve it. Also, when I print binaryform it prints like the way below. Shouldn't it print reversely I mean first 1 and then 11 and then 111 .......10111.
# Python program to convert decimal number into binary number using recursive function
def binary(n, binaryform,i):
if n >= 1:
digit= n % 2
binaryform += digit*i
#print(binaryform)
i*=10
binary(n/2, binaryform, i)
print("xxx", binaryform)
return binaryform
dec = int(input("Enter an integer: "))# Take decimal number from user
aa = binary(dec, 0, 1)
print(aa)
OUTPUT:
Enter an integer: 23
('xxx', 10111)
('xxx', 111)
('xxx', 111)
('xxx', 11)
('xxx', 1)
1
Convert an integer number to a binary string. The result is a valid Python expression. If x is not a Python int object, it has to define an index() method that returns an integer. Python Documentation
You're writing a recursive function - that's fun and good.
Now consider what you do: you check the LOW part of the number. If the LOW part of the number is 1 or 0 you do something, but then you shrink the number and recurse, checking a new LOW part, but the new low part came from a HIGHER part originally.
So whatever you determine should actually be at the end of the string, not the beginning, when you come back from recursion.
That's your question about printing reversely, I think. And yes, except you should just assemble it reversely, and then print it normally.
Also, you're actually trying to construct a decimal integer that will print out as if it were a binary number. It would be a lot simpler if you just constructed a string: "10101".
If you just want to convert integer to binary, try this code here . It looks like this :
def binary(n):
b = ''
while n > 0:
b = str(n % 2) + b
n >>= 1
print(b)
binary(10)
I'd simply like to convert a base-2 binary number string into an int, something like this:
>>> '11111111'.fromBinaryToInt()
255
Is there a way to do this in Python?
You use the built-in int() function, and pass it the base of the input number, i.e. 2 for a binary number:
>>> int('11111111', 2)
255
Here is documentation for Python 2, and for Python 3.
Just type 0b11111111 in python interactive interface:
>>> 0b11111111
255
Another way to do this is by using the bitstring module:
>>> from bitstring import BitArray
>>> b = BitArray(bin='11111111')
>>> b.uint
255
Note that the unsigned integer (uint) is different from the signed integer (int):
>>> b.int
-1
Your question is really asking for the unsigned integer representation; this is an important distinction.
The bitstring module isn't a requirement, but it has lots of performant methods for turning input into and from bits into other forms, as well as manipulating them.
Using int with base is the right way to go. I used to do this before I found int takes base also. It is basically a reduce applied on a list comprehension of the primitive way of converting binary to decimal ( e.g. 110 = 2**0 * 0 + 2 ** 1 * 1 + 2 ** 2 * 1)
add = lambda x,y : x + y
reduce(add, [int(x) * 2 ** y for x, y in zip(list(binstr), range(len(binstr) - 1, -1, -1))])
If you wanna know what is happening behind the scene, then here you go.
class Binary():
def __init__(self, binNumber):
self._binNumber = binNumber
self._binNumber = self._binNumber[::-1]
self._binNumber = list(self._binNumber)
self._x = [1]
self._count = 1
self._change = 2
self._amount = 0
print(self._ToNumber(self._binNumber))
def _ToNumber(self, number):
self._number = number
for i in range (1, len (self._number)):
self._total = self._count * self._change
self._count = self._total
self._x.append(self._count)
self._deep = zip(self._number, self._x)
for self._k, self._v in self._deep:
if self._k == '1':
self._amount += self._v
return self._amount
mo = Binary('101111110')
Here's another concise way to do it not mentioned in any of the above answers:
>>> eval('0b' + '11111111')
255
Admittedly, it's probably not very fast, and it's a very very bad idea if the string is coming from something you don't have control over that could be malicious (such as user input), but for completeness' sake, it does work.
A recursive Python implementation:
def int2bin(n):
return int2bin(n >> 1) + [n & 1] if n > 1 else [1]
If you are using python3.6 or later you can use f-string to do the
conversion:
Binary to decimal:
>>> print(f'{0b1011010:#0}')
90
>>> bin_2_decimal = int(f'{0b1011010:#0}')
>>> bin_2_decimal
90
binary to octal hexa and etc.
>>> f'{0b1011010:#o}'
'0o132' # octal
>>> f'{0b1011010:#x}'
'0x5a' # hexadecimal
>>> f'{0b1011010:#0}'
'90' # decimal
Pay attention to 2 piece of information separated by colon.
In this way, you can convert between {binary, octal, hexadecimal, decimal} to {binary, octal, hexadecimal, decimal} by changing right side of colon[:]
:#b -> converts to binary
:#o -> converts to octal
:#x -> converts to hexadecimal
:#0 -> converts to decimal as above example
Try changing left side of colon to have octal/hexadecimal/decimal.
For large matrix (10**5 rows and up) it is better to use a vectorized matmult. Pass in all rows and cols in one shot. It is extremely fast. There is no looping in python here. I originally designed it for converting many binary columns like 0/1 for like 10 different genre columns in MovieLens into a single integer for each example row.
def BitsToIntAFast(bits):
m,n = bits.shape
a = 2**np.arange(n)[::-1] # -1 reverses array of powers of 2 of same length as bits
return bits # a
For the record to go back and forth in basic python3:
a = 10
bin(a)
# '0b1010'
int(bin(a), 2)
# 10
eval(bin(a))
# 10
The task is to search every power of two below 2^10000, returning the index of the first power in which a string is contained. For example if the given string to search for is "7" the program will output 15, as 2^15 is the first power to contain 7 in it.
I have approached this with a brute force attempt which times out on ~70% of test cases.
for i in range(1,9999):
if search in str(2**i):
print i
break
How would one approach this with a time limit of 5 seconds?
Try not to compute 2^i at each step.
pow = 1
for i in xrange(1,9999):
if search in str(pow):
print i
break
pow *= 2
You can compute it as you go along. This should save a lot of computation time.
Using xrange will prevent a list from being built, but that will probably not make much of a difference here.
in is probably implemented as a quadratic string search algorithm. It may (or may not, you'd have to test) be more efficient to use something like KMP for string searching.
A faster approach could be computing the numbers directly in decimal
def double(x):
carry = 0
for i, v in enumerate(x):
d = v*2 + carry
if d > 99999999:
x[i] = d - 100000000
carry = 1
else:
x[i] = d
carry = 0
if carry:
x.append(carry)
Then the search function can become
def p2find(s):
x = [1]
for y in xrange(10000):
if s in str(x[-1])+"".join(("00000000"+str(y))[-8:]
for y in x[::-1][1:]):
return y
double(x)
return None
Note also that the digits of all powers of two up to 2^10000 are just 15 millions, and searching the static data is much faster. If the program must not be restarted each time then
def p2find(s, digits = []):
if len(digits) == 0:
# This precomputation happens only ONCE
p = 1
for k in xrange(10000):
digits.append(str(p))
p *= 2
for i, v in enumerate(digits):
if s in v: return i
return None
With this approach the first check will take some time, next ones will be very very fast.
Compute every power of two and build a suffix tree using each string. This is linear time in the size of all the strings. Now, the lookups are basically linear time in the length of each lookup string.
I don't think you can beat this for computational complexity.
There are only 10000 numbers. You don't need any complex algorithms. Simply calculated them in advance and do search. This should take merely 1 or 2 seconds.
powers_of_2 = [str(1<<i) for i in range(10000)]
def search(s):
for i in range(len(powers_of_2)):
if s in powers_of_2[i]:
return i
Try this
twos = []
twoslen = []
two = 1
for i in xrange(10000):
twos.append(two)
twoslen.append(len(str(two)))
two *= 2
tens = []
ten = 1
for i in xrange(len(str(two))):
tens.append(ten)
ten *= 10
s = raw_input()
l = len(s)
n = int(s)
for i in xrange(len(twos)):
for j in xrange(twoslen[i]):
k = twos[i] / tens[j]
if k < n: continue
if (k - n) % tens[l] == 0:
print i
exit()
The idea is to precompute every power of 2, 10 and and also to precompute the number of digits for every power of 2. In this way the problem is reduces to finding the minimum i for which there exist a j such that after removing the last j digits from 2 ** i you obtain a number which ends with n or expressed as a formula (2 ** i / 10 ** j - n) % 10 ** len(str(n)) == 0.
A big problem here is that converting a binary integer to decimal notation takes time quadratic in the number of bits (at least in the straightforward way Python does it). It's actually faster to fake your own decimal arithmetic, as #6502 did in his answer.
But it's very much faster to let Python's decimal module do it - at least under Python 3.3.2 (I don't know how much C acceleration is built in to Python decimal versions before that). Here's code:
class S:
def __init__(self):
import decimal
decimal.getcontext().prec = 4000 # way more than enough for 2**10000
p2 = decimal.Decimal(1)
full = []
for i in range(10000):
s = "%s<%s>" % (p2, i)
##assert s == "%s<%s>" % (str(2**i), i)
full.append(s)
p2 *= 2
self.full = "".join(full)
def find(self, s):
import re
pat = s + "[^<>]*<(\d+)>"
m = re.search(pat, self.full)
if m:
return int(m.group(1))
else:
print(s, "not found!")
and sample usage:
>>> s = S()
>>> s.find("1")
0
>>> s.find("2")
1
>>> s.find("3")
5
>>> s.find("65")
16
>>> s.find("7")
15
>>> s.find("00000")
1491
>>> s.find("666")
157
>>> s.find("666666")
2269
>>> s.find("66666666")
66666666 not found!
s.full is a string with a bit over 15 million characters. It looks like this:
>>> print(s.full[:20], "...", s.full[-20:])
1<0>2<1>4<2>8<3>16<4 ... 52396298354688<9999>
So the string contains each power of 2, with the exponent following a power enclosed in angle brackets. The find() method constructs a regular expression to search for the desired substring, then look ahead to find the power.
Playing around with this, I'm convinced that just about any way of searching is "fast enough". It's getting the decimal representations of the large powers that sucks up the vast bulk of the time. And the decimal module solves that one.
This is a rather difficult challenge for me as I am new to Python. How would I write a program in python based off this sequence function:
http://oeis.org/A063655
and does the following:
It asks for the value of the sequence and returns the corresponding number. For example, the number corresponding to the 10th value of the sequence is 7. I'd like to be able to do this for values over 300,000,000.
So, the final product would look like this:
Enter a value: 4
[7]
Any ideas where to start? I have a framework to generate sequences where (x) would be to put a mathematical equation or numbers, but I'm not exactly sure how to go from here or how to implement the "Enter a value" portion:
import math
def my_deltas():
while True:
yield (x)
yield (x)
def numbers(start, deltas, max):
i=start
while i<=max:
yield i
i+=next(deltas)
print(','.join(str(i) for i in numbers((x), my_deltas(),(x))))
If you're looking to have your computer keep track of over 300,000,000 elements of a sequence, if each is a 4 byte integer, you'll need at least 300,000,000 * 4bytes, or over 1.1GB of space to store all the values. I assume generating the sequence would also take a really long time, so generating the whole sequence again each time the user wants a value is not quite optimal either. I am a little confused about how you are trying to approach this exactly.
To get a value from the user is simple: you can use val = input("What is your value? ") where val is the variable you store it in.
EDIT:
It seems like a quick and simple approach would be this way, with a reasonable number of steps for each value (unless the value is prime...but lets keep the concept simple for now): You'd need the integer less than or equal to the square root of n (start_int = n ** .5), and from there you test each integer below to see if it divides n, first converting start_int to an integer with start_int = int(start_int) (which gives you the floor of start_int), like so: while (n % start_int) != 0: start_int = start_int - 1, decrement by one, and then set b = start_int. Something similar to find d, but you'll have to figure that part out. Note that % is the modulus operator (if you don't know what that is, you can read up on it, google: 'modulus python'), and ** is exponentiation. You can then return a value with the return statement. Your function would look something like this (lines starting with # are comments and python skips over them):
def find_number(value):
#using value instead of n
start_int = value ** .5
start_int = int(start_int)
while (n % start_int) != 0:
#same thing as start_int = start_int - 1
start_int -= 1
b = start_int
#...more code here
semiperimeter = b + d
return semiperimeter
#Let's use this function now!
#store
my_val = input("Enter a value: ")
my_number = find_number(my_val)
print my_number
There are many introductory guides to Python, and I would suggest you go through one first before tackling implementing a problem like this. If you already know how to program in another language you can just skim a guide to Python's syntax.
Don't forget to choose this answer if it helped!
from math import sqrt, floor
def A063655(n):
for i in range(floor(sqrt(n)), 0, -1):
j = floor(n / i)
if i * j == n:
return i + j
if __name__ == '__main__':
my_value = int(input("Enter a value: "))
my_number = A063655(my_value)
print(my_number)
USAGE
> python3 test.py
Enter a value: 10
7
> python3 test.py
Enter a value: 350000
1185
>