We Keep Coding

list of floats in python [duplicate]

How do I iterate between 0 and 1 by a step of 0.1?
This says that the step argument cannot be zero:
for i in range(0, 1, 0.1):
print(i)

Rather than using a decimal step directly, it's much safer to express this in terms of how many points you want. Otherwise, floating-point rounding error is likely to give you a wrong result.
Use the linspace function from the NumPy library (which isn't part of the standard library but is relatively easy to obtain). linspace takes a number of points to return, and also lets you specify whether or not to include the right endpoint:
>>> np.linspace(0,1,11)
array([ 0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1. ])
>>> np.linspace(0,1,10,endpoint=False)
array([ 0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9])
If you really want to use a floating-point step value, use numpy.arange:
>>> import numpy as np
>>> np.arange(0.0, 1.0, 0.1)
array([ 0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9])
Floating-point rounding error will cause problems, though. Here's a simple case where rounding error causes arange to produce a length-4 array when it should only produce 3 numbers:
>>> numpy.arange(1, 1.3, 0.1)
array([1. , 1.1, 1.2, 1.3])

range() can only do integers, not floating point.
Use a list comprehension instead to obtain a list of steps:
[x * 0.1 for x in range(0, 10)]
More generally, a generator comprehension minimizes memory allocations:
xs = (x * 0.1 for x in range(0, 10))
for x in xs:
print(x)

Building on 'xrange([start], stop[, step])', you can define a generator that accepts and produces any type you choose (stick to types supporting + and <):
>>> def drange(start, stop, step):
... r = start
... while r < stop:
... yield r
... r += step
...
>>> i0=drange(0.0, 1.0, 0.1)
>>> ["%g" % x for x in i0]
['0', '0.1', '0.2', '0.3', '0.4', '0.5', '0.6', '0.7', '0.8', '0.9', '1']
>>>

Increase the magnitude of i for the loop and then reduce it when you need it.
for i * 100 in range(0, 100, 10):
print i / 100.0
EDIT: I honestly cannot remember why I thought that would work syntactically
for i in range(0, 11, 1):
print i / 10.0
That should have the desired output.

NumPy is a bit overkill, I think.
[p/10 for p in range(0, 10)]
[0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9]
Generally speaking, to do a step-by-1/x up to y you would do
x=100
y=2
[p/x for p in range(0, int(x*y))]
[0.0, 0.01, 0.02, 0.03, ..., 1.97, 1.98, 1.99]
(1/x produced less rounding noise when I tested).

scipy has a built in function arange which generalizes Python's range() constructor to satisfy your requirement of float handling.
from scipy import arange

Similar to R's seq function, this one returns a sequence in any order given the correct step value. The last value is equal to the stop value.
def seq(start, stop, step=1):
n = int(round((stop - start)/float(step)))
if n > 1:
return([start + step*i for i in range(n+1)])
elif n == 1:
return([start])
else:
return([])
Results
seq(1, 5, 0.5)
[1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0]
seq(10, 0, -1)
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
seq(10, 0, -2)
[10, 8, 6, 4, 2, 0]
seq(1, 1)
[ 1 ]

The range() built-in function returns a sequence of integer values, I'm afraid, so you can't use it to do a decimal step.
I'd say just use a while loop:
i = 0.0
while i <= 1.0:
print i
i += 0.1
If you're curious, Python is converting your 0.1 to 0, which is why it's telling you the argument can't be zero.

Here's a solution using itertools:
import itertools
def seq(start, end, step):
if step == 0:
raise ValueError("step must not be 0")
sample_count = int(abs(end - start) / step)
return itertools.islice(itertools.count(start, step), sample_count)
Usage Example:
for i in seq(0, 1, 0.1):
print(i)

[x * 0.1 for x in range(0, 10)]
in Python 2.7x gives you the result of:
[0.0, 0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6000000000000001, 0.7000000000000001, 0.8, 0.9]
but if you use:
[ round(x * 0.1, 1) for x in range(0, 10)]
gives you the desired:
[0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9]

import numpy as np
for i in np.arange(0, 1, 0.1):
print i

Best Solution: no rounding error
>>> step = .1
>>> N = 10 # number of data points
>>> [ x / pow(step, -1) for x in range(0, N + 1) ]
[0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]
Or, for a set range instead of set data points (e.g. continuous function), use:
>>> step = .1
>>> rnge = 1 # NOTE range = 1, i.e. span of data points
>>> N = int(rnge / step
>>> [ x / pow(step,-1) for x in range(0, N + 1) ]
[0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]
To implement a function: replace x / pow(step, -1) with f( x / pow(step, -1) ), and define f.
For example:
>>> import math
>>> def f(x):
return math.sin(x)
>>> step = .1
>>> rnge = 1 # NOTE range = 1, i.e. span of data points
>>> N = int(rnge / step)
>>> [ f( x / pow(step,-1) ) for x in range(0, N + 1) ]
[0.0, 0.09983341664682815, 0.19866933079506122, 0.29552020666133955, 0.3894183423086505,
0.479425538604203, 0.5646424733950354, 0.644217687237691, 0.7173560908995228,
0.7833269096274834, 0.8414709848078965]

And if you do this often, you might want to save the generated list r
r=map(lambda x: x/10.0,range(0,10))
for i in r:
print i

more_itertools is a third-party library that implements a numeric_range tool:
import more_itertools as mit
for x in mit.numeric_range(0, 1, 0.1):
print("{:.1f}".format(x))
Output
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
This tool also works for Decimal and Fraction.

My versions use the original range function to create multiplicative indices for the shift. This allows same syntax to the original range function.
I have made two versions, one using float, and one using Decimal, because I found that in some cases I wanted to avoid the roundoff drift introduced by the floating point arithmetic.
It is consistent with empty set results as in range/xrange.
Passing only a single numeric value to either function will return the standard range output to the integer ceiling value of the input parameter (so if you gave it 5.5, it would return range(6).)
Edit: the code below is now available as package on pypi: Franges
## frange.py
from math import ceil
# find best range function available to version (2.7.x / 3.x.x)
try:
_xrange = xrange
except NameError:
_xrange = range
def frange(start, stop = None, step = 1):
"""frange generates a set of floating point values over the
range [start, stop) with step size step
frange([start,] stop [, step ])"""
if stop is None:
for x in _xrange(int(ceil(start))):
yield x
else:
# create a generator expression for the index values
indices = (i for i in _xrange(0, int((stop-start)/step)))
# yield results
for i in indices:
yield start + step*i
## drange.py
import decimal
from math import ceil
# find best range function available to version (2.7.x / 3.x.x)
try:
_xrange = xrange
except NameError:
_xrange = range
def drange(start, stop = None, step = 1, precision = None):
"""drange generates a set of Decimal values over the
range [start, stop) with step size step
drange([start,] stop, [step [,precision]])"""
if stop is None:
for x in _xrange(int(ceil(start))):
yield x
else:
# find precision
if precision is not None:
decimal.getcontext().prec = precision
# convert values to decimals
start = decimal.Decimal(start)
stop = decimal.Decimal(stop)
step = decimal.Decimal(step)
# create a generator expression for the index values
indices = (
i for i in _xrange(
0,
((stop-start)/step).to_integral_value()
)
)
# yield results
for i in indices:
yield float(start + step*i)
## testranges.py
import frange
import drange
list(frange.frange(0, 2, 0.5)) # [0.0, 0.5, 1.0, 1.5]
list(drange.drange(0, 2, 0.5, precision = 6)) # [0.0, 0.5, 1.0, 1.5]
list(frange.frange(3)) # [0, 1, 2]
list(frange.frange(3.5)) # [0, 1, 2, 3]
list(frange.frange(0,10, -1)) # []

Lots of the solutions here still had floating point errors in Python 3.6 and didnt do exactly what I personally needed.
Function below takes integers or floats, doesnt require imports and doesnt return floating point errors.
def frange(x, y, step):
if int(x + y + step) == (x + y + step):
r = list(range(int(x), int(y), int(step)))
else:
f = 10 ** (len(str(step)) - str(step).find('.') - 1)
rf = list(range(int(x * f), int(y * f), int(step * f)))
r = [i / f for i in rf]
return r

Suprised no-one has yet mentioned the recommended solution in the Python 3 docs:
See also:
The linspace recipe shows how to implement a lazy version of range that suitable for floating point applications.
Once defined, the recipe is easy to use and does not require numpy or any other external libraries, but functions like numpy.linspace(). Note that rather than a step argument, the third num argument specifies the number of desired values, for example:
print(linspace(0, 10, 5))
# linspace(0, 10, 5)
print(list(linspace(0, 10, 5)))
# [0.0, 2.5, 5.0, 7.5, 10]
I quote a modified version of the full Python 3 recipe from Andrew Barnert below:
import collections.abc
import numbers
class linspace(collections.abc.Sequence):
"""linspace(start, stop, num) -> linspace object
Return a virtual sequence of num numbers from start to stop (inclusive).
If you need a half-open range, use linspace(start, stop, num+1)[:-1].
"""
def __init__(self, start, stop, num):
if not isinstance(num, numbers.Integral) or num <= 1:
raise ValueError('num must be an integer > 1')
self.start, self.stop, self.num = start, stop, num
self.step = (stop-start)/(num-1)
def __len__(self):
return self.num
def __getitem__(self, i):
if isinstance(i, slice):
return [self[x] for x in range(*i.indices(len(self)))]
if i < 0:
i = self.num + i
if i >= self.num:
raise IndexError('linspace object index out of range')
if i == self.num-1:
return self.stop
return self.start + i*self.step
def __repr__(self):
return '{}({}, {}, {})'.format(type(self).__name__,
self.start, self.stop, self.num)
def __eq__(self, other):
if not isinstance(other, linspace):
return False
return ((self.start, self.stop, self.num) ==
(other.start, other.stop, other.num))
def __ne__(self, other):
return not self==other
def __hash__(self):
return hash((type(self), self.start, self.stop, self.num))

This is my solution to get ranges with float steps.
Using this function it's not necessary to import numpy, nor install it.
I'm pretty sure that it could be improved and optimized. Feel free to do it and post it here.
from __future__ import division
from math import log
def xfrange(start, stop, step):
old_start = start #backup this value
digits = int(round(log(10000, 10)))+1 #get number of digits
magnitude = 10**digits
stop = int(magnitude * stop) #convert from
step = int(magnitude * step) #0.1 to 10 (e.g.)
if start == 0:
start = 10**(digits-1)
else:
start = 10**(digits)*start
data = [] #create array
#calc number of iterations
end_loop = int((stop-start)//step)
if old_start == 0:
end_loop += 1
acc = start
for i in xrange(0, end_loop):
data.append(acc/magnitude)
acc += step
return data
print xfrange(1, 2.1, 0.1)
print xfrange(0, 1.1, 0.1)
print xfrange(-1, 0.1, 0.1)
The output is:
[1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0]
[0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1]
[-1.0, -0.9, -0.8, -0.7, -0.6, -0.5, -0.4, -0.3, -0.2, -0.1, 0.0]

For completeness of boutique, a functional solution:
def frange(a,b,s):
return [] if s > 0 and a > b or s < 0 and a < b or s==0 else [a]+frange(a+s,b,s)

You can use this function:
def frange(start,end,step):
return map(lambda x: x*step, range(int(start*1./step),int(end*1./step)))

It can be done using Numpy library. arange() function allows steps in float. But, it returns a numpy array which can be converted to list using tolist() for our convenience.
for i in np.arange(0, 1, 0.1).tolist():
print i

start and stop are inclusive rather than one or the other (usually stop is excluded) and without imports, and using generators
def rangef(start, stop, step, fround=5):
"""
Yields sequence of numbers from start (inclusive) to stop (inclusive)
by step (increment) with rounding set to n digits.
:param start: start of sequence
:param stop: end of sequence
:param step: int or float increment (e.g. 1 or 0.001)
:param fround: float rounding, n decimal places
:return:
"""
try:
i = 0
while stop >= start and step > 0:
if i==0:
yield start
elif start >= stop:
yield stop
elif start < stop:
if start == 0:
yield 0
if start != 0:
yield start
i += 1
start += step
start = round(start, fround)
else:
pass
except TypeError as e:
yield "type-error({})".format(e)
else:
pass
# passing
print(list(rangef(-100.0,10.0,1)))
print(list(rangef(-100,0,0.5)))
print(list(rangef(-1,1,0.2)))
print(list(rangef(-1,1,0.1)))
print(list(rangef(-1,1,0.05)))
print(list(rangef(-1,1,0.02)))
print(list(rangef(-1,1,0.01)))
print(list(rangef(-1,1,0.005)))
# failing: type-error:
print(list(rangef("1","10","1")))
print(list(rangef(1,10,"1")))
Python 3.6.2 (v3.6.2:5fd33b5, Jul 8 2017, 04:57:36) [MSC v.1900 64
bit (AMD64)]

I know I'm late to the party here, but here's a trivial generator solution that's working in 3.6:
def floatRange(*args):
start, step = 0, 1
if len(args) == 1:
stop = args[0]
elif len(args) == 2:
start, stop = args[0], args[1]
elif len(args) == 3:
start, stop, step = args[0], args[1], args[2]
else:
raise TypeError("floatRange accepts 1, 2, or 3 arguments. ({0} given)".format(len(args)))
for num in start, step, stop:
if not isinstance(num, (int, float)):
raise TypeError("floatRange only accepts float and integer arguments. ({0} : {1} given)".format(type(num), str(num)))
for x in range(int((stop-start)/step)):
yield start + (x * step)
return
then you can call it just like the original range()... there's no error handling, but let me know if there is an error that can be reasonably caught, and I'll update. or you can update it. this is StackOverflow.

To counter the float precision issues, you could use the Decimal module.
This demands an extra effort of converting to Decimal from int or float while writing the code, but you can instead pass str and modify the function if that sort of convenience is indeed necessary.
from decimal import Decimal
def decimal_range(*args):
zero, one = Decimal('0'), Decimal('1')
if len(args) == 1:
start, stop, step = zero, args[0], one
elif len(args) == 2:
start, stop, step = args + (one,)
elif len(args) == 3:
start, stop, step = args
else:
raise ValueError('Expected 1 or 2 arguments, got %s' % len(args))
if not all([type(arg) == Decimal for arg in (start, stop, step)]):
raise ValueError('Arguments must be passed as <type: Decimal>')
# neglect bad cases
if (start == stop) or (start > stop and step >= zero) or \
(start < stop and step <= zero):
return []
current = start
while abs(current) < abs(stop):
yield current
current += step
Sample outputs -
from decimal import Decimal as D
list(decimal_range(D('2')))
# [Decimal('0'), Decimal('1')]
list(decimal_range(D('2'), D('4.5')))
# [Decimal('2'), Decimal('3'), Decimal('4')]
list(decimal_range(D('2'), D('4.5'), D('0.5')))
# [Decimal('2'), Decimal('2.5'), Decimal('3.0'), Decimal('3.5'), Decimal('4.0')]
list(decimal_range(D('2'), D('4.5'), D('-0.5')))
# []
list(decimal_range(D('2'), D('-4.5'), D('-0.5')))
# [Decimal('2'),
# Decimal('1.5'),
# Decimal('1.0'),
# Decimal('0.5'),
# Decimal('0.0'),
# Decimal('-0.5'),
# Decimal('-1.0'),
# Decimal('-1.5'),
# Decimal('-2.0'),
# Decimal('-2.5'),
# Decimal('-3.0'),
# Decimal('-3.5'),
# Decimal('-4.0')]

Add auto-correction for the possibility of an incorrect sign on step:
def frange(start,step,stop):
step *= 2*((stop>start)^(step<0))-1
return [start+i*step for i in range(int((stop-start)/step))]

My solution:
def seq(start, stop, step=1, digit=0):
x = float(start)
v = []
while x <= stop:
v.append(round(x,digit))
x += step
return v

Here is my solution which works fine with float_range(-1, 0, 0.01) and works without floating point representation errors. It is not very fast, but works fine:
from decimal import Decimal
def get_multiplier(_from, _to, step):
digits = []
for number in [_from, _to, step]:
pre = Decimal(str(number)) % 1
digit = len(str(pre)) - 2
digits.append(digit)
max_digits = max(digits)
return float(10 ** (max_digits))
def float_range(_from, _to, step, include=False):
"""Generates a range list of floating point values over the Range [start, stop]
with step size step
include=True - allows to include right value to if possible
!! Works fine with floating point representation !!
"""
mult = get_multiplier(_from, _to, step)
# print mult
int_from = int(round(_from * mult))
int_to = int(round(_to * mult))
int_step = int(round(step * mult))
# print int_from,int_to,int_step
if include:
result = range(int_from, int_to + int_step, int_step)
result = [r for r in result if r <= int_to]
else:
result = range(int_from, int_to, int_step)
# print result
float_result = [r / mult for r in result]
return float_result
print float_range(-1, 0, 0.01,include=False)
assert float_range(1.01, 2.06, 5.05 % 1, True) ==\
[1.01, 1.06, 1.11, 1.16, 1.21, 1.26, 1.31, 1.36, 1.41, 1.46, 1.51, 1.56, 1.61, 1.66, 1.71, 1.76, 1.81, 1.86, 1.91, 1.96, 2.01, 2.06]
assert float_range(1.01, 2.06, 5.05 % 1, False)==\
[1.01, 1.06, 1.11, 1.16, 1.21, 1.26, 1.31, 1.36, 1.41, 1.46, 1.51, 1.56, 1.61, 1.66, 1.71, 1.76, 1.81, 1.86, 1.91, 1.96, 2.01]

I am only a beginner, but I had the same problem, when simulating some calculations. Here is how I attempted to work this out, which seems to be working with decimal steps.
I am also quite lazy and so I found it hard to write my own range function.
Basically what I did is changed my xrange(0.0, 1.0, 0.01) to xrange(0, 100, 1) and used the division by 100.0 inside the loop.
I was also concerned, if there will be rounding mistakes. So I decided to test, whether there are any. Now I heard, that if for example 0.01 from a calculation isn't exactly the float 0.01 comparing them should return False (if I am wrong, please let me know).
So I decided to test if my solution will work for my range by running a short test:
for d100 in xrange(0, 100, 1):
d = d100 / 100.0
fl = float("0.00"[:4 - len(str(d100))] + str(d100))
print d, "=", fl , d == fl
And it printed True for each.
Now, if I'm getting it totally wrong, please let me know.

The trick to avoid round-off problem is to use a separate number to move through the range, that starts and half the step ahead of start.
# floating point range
def frange(a, b, stp=1.0):
i = a+stp/2.0
while i<b:
yield a
a += stp
i += stp
Alternatively, numpy.arange can be used.

My answer is similar to others using map(), without need of NumPy, and without using lambda (though you could). To get a list of float values from 0.0 to t_max in steps of dt:
def xdt(n):
return dt*float(n)
tlist = map(xdt, range(int(t_max/dt)+1))

Elements in list greater than or equal to elements in other list (without for loop?)

I have a list containing 1,000,000 elements (numbers) called x and I would like to count how many of them are equal to or above [0.5,0.55,0.60,...,1]. Is there a way to do it without a for loop?
Right now I have the following the code, which works for a specific value of the [0.5,...1] interval, let's say 0.5 and assigns it to the count variable
count=len([i for i in x if i >= 0.5])
EDIT: Basically what I want to avoid is doing this... if possible?
obs=[]
alpha = [0.5,0.55,0.6,0.65,0.7,0.75,0.8,0.85,0.9,0.95,1]
for a in alpha:
count= len([i for i in x if i >= a])
obs.append(count)
Thanks in advance
Best, Mikael

I don't think it's possible without loop, but you can sort the array x and then you can use bisect module (doc) to locate insertion point (index).
For example:
x = [0.341, 0.423, 0.678, 0.999, 0.523, 0.751, 0.7]
alpha = [0.5,0.55,0.6,0.65,0.7,0.75,0.8,0.85,0.9,0.95,1]
x = sorted(x)
import bisect
obs = [len(x) - bisect.bisect_left(x, a) for a in alpha]
print(obs)
Will print:
[5, 4, 4, 4, 3, 2, 1, 1, 1, 1, 0]
Note:
sorted() has complexity n log(n) and bisect_left() log(n)

You can use numpy and boolean indexing:
>>> import numpy as np
>>> a = np.array(list(range(100)))
>>> a[a>=50].size
50

Even if you are not using for loop, internal methods use them. But iterates them efficiently.
you can use below function without for loop from your end.
x = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9]
l = list(filter(lambda _: _ > .5 , x))
print(l)

Based on comments, you're ok with using numpy, so use np.searchsorted to simply insert alpha into a sorted version of x. The indices will be your counts.
If you're ok with sorting x in-place:
x.sort()
counts = x.size - np.searchsorted(x, alpha)
If not,
counts = x.size - np.searchsorted(np.sort(x), alpha)
These counts assume that you want x < alpha. To get <= add the keyword side='right':
np.searchsorted(x, alpha, side='right')
PS
There are a couple of significant problems with the line
count = len([i for i in x if i >= 0.5])
First of all, you're creating a list of all the matching elements instead of just counting them. To count them do
count = sum(1 for i in x if i >= threshold)
Now the problem is that you are doing a linear pass through the entire array for each alpha, which is not necessary.
As I commented under #Andrej Kesely's answer, let's say we have N = len(x) and M = len(alpha). Your implementation is O(M * N) time complexity, while sorting gives you O((M + N) log N). For M << N (small alpha), your complexity is approximately O(N), which beats O(N log N). But for M ~= N, yours approaches O(N^2) vs my O(N log N).

EDIT: If you are using NumPy already, you can simply do this:
import numpy as np
# Make random data
np.random.seed(0)
x = np.random.binomial(n=20, p=0.5, size=1000000) / 20
bins = np.arange(0.55, 1.01, 0.05)
# One extra value for the upper bound of last bin
bins = np.append(bins, max(bins.max(), x.max()) + 1)
h, _ = np.histogram(x, bins)
result = np.cumsum(h)
print(result)
# [280645 354806 391658 406410 411048 412152 412356 412377 412378 412378]
If you are dealing with large arrays of numbers, you may considering using NumPy. But if you are using simple Python lists, you can do that for example like this:
def how_many_bigger(nums, mins):
# List of counts for each minimum
counts = [0] * len(mins)
# For each number
for n in nums:
# For each minimum
for i, m in enumerate(mins):
# Add 1 to the count if the number is greater than the current minimum
if n >= m:
counts[i] += 1
return counts
# Test
import random
# Make random data
random.seed(0)
nums = [random.random() for _ in range(1_000_000)]
# Make minimums
mins = [i / 100. for i in range(55, 101, 5)]
print(mins)
# [0.55, 0.6, 0.65, 0.7, 0.75, 0.8, 0.85, 0.9, 0.95, 1.0]
count = how_many_bigger(nums, mins)
print(count)
# [449771, 399555, 349543, 299687, 249605, 199774, 149945, 99928, 49670, 0]

Transformation matrix in Python

I have a scale in a 3D space from -10 to 10 with step of 2.5 ([-10,-7.5,-5,-2.5,0,2.5,5,7.5,10]).
I have a 3D point in this scale, I want mapped it in another 3D space from 0 to 8 with step of 1 ([0,1,2,3,4,5,6,7,8]).
How can I do it?
Thanks for your help

In this stack exachange post you find the mathematical formula for normalization within a range. The python implementation could be the following:
def normalize(values, curr_bounds, new_bounds):
return [(x - curr_bounds[0]) * (new_bounds[1] - new_bounds[0]) / (curr_bounds[1] - curr_bounds[0] + new_bounds[0]) for x in values]
Your request is not very clear, if your data needs to be mapped in a discrete range (rounded by step) then you can do this:
def normalize_step(values, curr_bounds, new_bounds, new_step):
return [round_step(new_step, (x - curr_bounds[0]) * (new_bounds[1] - new_bounds[0]) / (curr_bounds[1] - curr_bounds[0] + new_bounds[0])) for x in values]
def round_step(step, n):
return (n//step + 1) * step if n%step >= step/2 else n//step * step
For example, given the following data:
current_bounds = (-10, 10)
new_bounds = (0, 8)
step = 1
values = [-10, -2.5, 0, 7.5, 2.5]
normalize(values, current_bounds, new_bounds)
# [0.0, 3.0, 4.0, 7.0, 5.0]
normalize_step(values, current_bounds, new_bounds, step)
# [0.0, 3.0, 4.0, 7.0, 5.0]
Note: In this case the result is the same because your step=1, if we put the step=1.5 the results change:
normalize_step(values, current_bounds, new_bounds, 1.5)
# [0.0, 3.0, 4.5, 7.5, 4.5]

How to cluster continuous peak widths

I have a sample data on different prices of a single product and how many times they have been entered in the database. My objective is to determine the optimal price range/(s) for the product based on its frequency in the data set.
This is the price-frequency plot:
I am using Python 3.7.1 on Jupyter. I tried using peak_widths function from the signal library. I got the peak widths, their start points and end points. I made an if-else loop logic but I think I'm going wrong somewhere, I'm getting only a single cluster.
I used this to get peak widths
from scipy.signal import peak_widths
pkwdh_106 = signal.peak_widths(count_106, index_106, rel_height=1)
plt.plot(index_106, count_106[index_106], "o"); plt.plot(count_106)
plt.hlines(*pkwdh_106[1:])
plt.xlabel('Price')
plt.ylabel('Frequency')
This is the peaks' information
Width of each peak / maxima : [7. 5. 4. 7. 2.88888889]
Y level of each width : [1. 1. 1. 1. 2.]
Starting point of each width : [ 8. 15. 20. 36. 40.]
Ending point of each width : [15. 20. 24. 43. 42.88888889]
This is the loop that I tried but something is missing here.
sum_width = 0
name='cluster('+str(i)+')'
name=[]
start = len(start_x) # Length of starting array is 5
end = len(end_x) # Length of ending array is 5
for i in range(end-1):
# Iterating till 4 END elements
for j in range(start-1):
# Iterating till 4 START elements
if end_x[i]>=start_x[j+1]:
# If ending of 1st width is greater than or equal to starting of 2nd width
sum_width = sum_width + width[i]
# Add starting of 1st width length to the total sum variable
name.append(width[i])
# Adding the width lengths to an array
print('If loop total width - ', sum_width)
print('If loop ', name)
else:
sum_width = sum_width + width[i]
# If ending of 1st width is lesser than the starting of 2nd width
name.append(width[i])
# Add starting of 1st width length to the total sum variable
print('Else loop total width - ', sum_width)
print('Else loop ', name)
break
break
print(sum_width)
This is the actual output that I am getting -
If loop total width - 7.0
If loop [7.0]
7.0
If loop total width - 12.0
If loop [7.0, 5.0]
12.0
If loop total width - 16.0
If loop [7.0, 5.0, 4.0]
16.0
If loop total width - 23.0
If loop [7.0, 5.0, 4.0, 7.0]
23.0
I expect to get two clusters like -
[7.0, 5.0, 4.0] [7.0, 2.88888889]

Currently your code will append the width to the name list regardless whether the condition (end_x[i]>=start_x[j+1]) is true or false. It then directly breaks out of the second loop, making it redundant.
Some changes need to be made:
You only need to do one-pass through the end_x and start_x lists. So only one for-loop is needed.
You'll need to keep track of a separate list for each cluster.
start_x = [ 8., 15., 20., 36., 40.]
end_x = [15., 20., 24., 43., 42.88888889]
width = [7., 5., 4., 7., 2.88888889]
sum_width = 0
clusters = [[width[0]]] # initialise with first width
for i in range(1, len(end_x)):
print(f'iter {i}: (start={start_x[i]}, end={end_x[i]}); width {width[i]}')
if end_x[i-1] < start_x[i]:
print(f' doing a reset ({end_x[i-1]} < {start_x[i]})')
sum_width = 0
clusters.append([])
sum_width += width[i]
print(' total width:', sum_width)
clusters[-1].append(width[i]) # append width into last cluster
print(clusters)
# [[7.0, 5.0, 4.0], [7.0, 2.88888889]]
Here, we iterate from second element to the end of the list, comparing the previous end_x element with the current start_x element and make a new cluster if values don't overlap.

python numpy - is there a faster way to convolve?

I have a numpy array that is very large (1 million integers). I'm using np.convolve in order to find the "densest" area of that array. By "desnsest" area I mean the window of a fixed length that has the the highest numbers when the window is summed. Let me show you in code:
import numpy as np
example = np.array([0,0,0,1,1,1,1,1,1,1,0,1,1,1,1,0,0,0,1,0,0,1,1,0,1,0,0,0,0,0,1,0])
window_size = 10
density = np.convolve(example, np.ones([window_size]), mode='valid')
print(density)
# [7.0, 7.0, 8.0, 9.0, 9.0, 9.0, 8.0, 7.0, 6.0, 6.0, 5.0, 5.0, 5.0, 5.0, 4.0, 4.0, 4.0, 4.0, 4.0, 3.0, 3.0, 4.0, 3.0]
I can then use np.argmax(density) to get the starting index of the highest density area 3.
Anyway, with this example it runs fast. but when convolving over million element array and with a window size of 10,000 it takes 2 seconds to complete. if I choose a windows_size of 500,000 it takes 3 minutes to complete.
Is there a better way to sum over the array with a certain window size to speed this up? If I converted this into a pandas series instead could I perhaps use something there?
Thanks for your help!

Try using scipy.signal.convolve. It has the option to compute the convolution using the fast Fourier transform (FFT), which should be much faster for the array sizes that you mentioned.
Using an array example with length 1000000 and convolving it with an array of length 10000, np.convolve took about 1.45 seconds on my computer, and scipy.signal.convolve took 22.7 milliseconds.

cumsum = np.cumsum(np.insert(example, 0, 0))
density2 = cumsum[window_size:]-cumsum[:-window_size]
np.all(density2 == density)
True
(remove insertion if you can live without the first value...)

This is how you can use the built-in NumPy real FFT functions to convolve in 1 dimension:
import numpy, numpy.fft.fftpack_lite
def fftpack_lite_rfftb(buf, s):
n = len(buf)
m = (n - 1) * 2
temp = numpy.empty(m, buf.dtype)
numpy.divide(buf, m, temp[:n])
temp[n:m] = 0
return numpy.fft.fftpack_lite.rfftb(temp[:m], s)
def fftconvolve(x, y):
xn = x.shape[-1]
yn = y.shape[-1]
cn = xn + yn - (xn + yn > 0)
m = 1 << cn.bit_length()
s = numpy.fft.fftpack_lite.rffti(m) # Initialization; can be factored out for performance
xpad = numpy.pad(x, [(0, 0)] * (len(x.shape) - 1) + [(0, m - xn)], 'constant')
a = numpy.fft.fftpack_lite.rfftf(xpad, s) # Forward transform
ypad = numpy.pad(y, [(0, 0)] * (len(y.shape) - 1) + [(0, m - yn)], 'constant')
b = numpy.fft.fftpack_lite.rfftf(ypad, s) # Forward transform
numpy.multiply(a, b, b) # Spectral multiplication
c = fftpack_lite_rfftb(b, s) # Backward transform
return c[:cn]
# Verify convolution is correct
assert (lambda a, b: numpy.allclose(fftconvolve(a, b), numpy.convolve(a, b)))(numpy.random.randn(numpy.random.randint(1, 32)), numpy.random.randn(numpy.random.randint(1, 32)))
Bear in mind that this padding is inefficient for convolution of vectors with significantly different sizes (> 100%); you'll want to use a linear combination technique like overlap-add to do smaller convolution.