numpy arange for floating point start & stop - python-2.7

I want numpy to create a full list, given the parameters start, stop and increment, but ran into some troubles:
In[2]: import numpy as np
In[3]: np.arange(2.0, 2.4, 0.2)
Out[3]: array([ 2. , 2.2])
In[4]: np.arange(2.0, 2.6, 0.2)
Out[4]: array([ 2. , 2.2, 2.4, 2.6])
In[5]: np.arange(2.0, 2.8, 0.2)
Out[5]: array([ 2. , 2.2, 2.4, 2.6])
What I actually want is:
array([ 2. , 2.2, 2.4])
Now, I've learned that I should avoid the floating point data type if it comes down to fixed values. I know it would be better to multiply start/stop/increment by 100, but the problem is that I cannot tell, how many decimals the user is going to supply. Is there any way I can still do that with Float or is there a better way to solve this?
Edit:
It works now with the obvious solution of adding 0.0000001 to the end-value. But this looks horrible in my code...I'd hope to fix this nicely somehow

Could you specify which values the user is supposed to enter? For that kind of generation, I think linspace could be better as it includes the end parameter
EDIT: if the user enters start, end, and increment, just use linspace with num = int((end-start)/increment+1) if the exact value of the increment is not critical.
EDIT2:
adapt 1e-4 to the relative error you deem acceptable (you can even add it as a user-defined variable).
eps = 1e-4*(stop-start)
num = int((stop-start)/(incr-eps)+1)
np.linspace(start, stop,num)

this might seem a little longer but if you are keen on using np.arange this is how I worked it out:
decimal_places = decimal.Decimal(str(STEP)).as_tuple().exponent
power_10_multiplier = 10**-decimal_places
MIN = int(MIN*power_10_multiplier)
MAX = int(MAX*power_10_multiplier)
STEP = int(STEP*power_10_multiplier)
arr = np.arange(MIN, MAX + STEP, step=STEP)/power_10_multiplier

Related

Specifying the Mean and Variance in a Scipy Distribution Python 2.7

I need to randomly sample from some distribution eventually, so I need one that allows me to readily change the mean and variance. I'm looking at using distributions from the scipy.stats library, however, I'm having difficulty seeing how the parameters "loc" and "scale" relate to the quantites I'm interested in. I'd like to be able to do something like:
x = numpy.linspace(0,5,1000)
y = scipy.stats.maxwell(x, mean, variance)
But loc and scale seem to be the only other arguments that function takes.
Can anyone specify the relationship those quantities must have to mean and variance, or suggest a better library to use?
Well, I don't have python 2.7, so answer would be for python 3.6, but it should work, it is a scipy after all.
Basically, you have to extract scale and loc parameters from given μ and σ. Here are two simple functions to do that, plus some sampling to prove we're getting right values. Basically, first printed line is what you want, and third line is result of sampling, should be roughly be the same. Second line is scale and loccomputed from μ and σ. Play with the numbers, see how it is going
import numpy as np
from scipy.stats import maxwell
def get_scale_from_sigma(sigma):
"""Compute scale from sigma based upon http://mathworld.wolfram.com/MaxwellDistribution.html"""
a2 = np.pi*sigma / (3.0*np.pi - 8.0)
return np.sqrt(a2)
def get_loc_from_mu_sigma(mu, sigma):
"""Compute loc from mu/sigma based upon http://mathworld.wolfram.com/MaxwellDistribution.html"""
scale = get_scale_from_sigma(sigma)
loc = mu - 2.0 * scale * np.sqrt(2.0 / np.pi)
return loc
sigma = 1.0
mu = 2.0 * get_scale_from_sigma(sigma) * np.sqrt(2.0 / np.pi) # + 3.0 as shift, for exampl
print(mu, sigma)
scale = get_scale_from_sigma(sigma)
loc = get_loc_from_mu_sigma(mu, sigma)
print(scale, loc)
q = maxwell.rvs(size=10000, scale = scale, loc = loc)
print(np.mean(q), np.std(q))

How can I remove indices of non-max values that correspond to duplicate values of separate list from both lists?

I have two lists, the first of which represents times of observation and the second of which represents the observed values at those times. I am trying to find the maximum observed value and the corresponding time given a rolling window of various length. For example-sake, here are the two lists.
# observed values
linspeed = [280.0, 275.0, 300.0, 475.2, 360.1, 400.9, 215.3, 323.8, 289.7]
# times that correspond to observed values
time_count = [4.0, 6.0, 8.0, 8.0, 10.0, 10.0, 10.0, 14.0, 16.0]
# actual dataset is of size ~ 11,000
The missing times (ex: 3.0) correspond to an observed value of zero, whereas duplicate times correspond to multiple observations to the floored time. Since my window will be rolling over the time_count (ex: max value in first 2 hours, next 2 hours, 2 hours after that; max value in first 4 hours, next 4 hours, ...), I plan to use an array-reshaping routine. However, it's important to set up everything properly before, which entails finding the maximum value given duplicate times. To solve this problem, I tried the code just below.
def list_duplicates(data_list):
seen = set()
seen_add = seen.add
seen_twice = set(x for x in data_list if x in seen or seen_add(x))
return list(seen_twice)
# check for duplicate values
dups = list_duplicates(time_count)
print(dups)
>> [8.0, 10.0]
# get index of duplicates
for dup in dups:
print(time_count.index(dup))
>> 2
>> 4
When checking for the index of the duplicates, it appears that this code will only return the index of the first occurrence of the duplicate value. I also tried using OrderedDict via module collections for reasons concerning code efficiency/speed, but dictionaries have a similar problem. Given duplicate keys for non-duplicate observation values, the first instance of the duplicate key and corresponding observation value is kept while all others are dropped from the dict. Per this SO post, my second attempt is just below.
for dup in dups:
indexes = [i for i,x in enumerate(time_count) if x == dup]
print(indexes)
>> [4, 5, 6] # indices correspond to duplicate time 10s but not duplicate time 8s
I should be getting [2,3] for time in time_count = 8.0 and [4,5,6] for time in time_count = 10.0. From the duplicate time_counts, 475.2 is the max linspeed that corresponds to duplicate time_count 8.0 and 400.9 is the max linspeed that corresponds to duplicate time_count 10.0, meaning that the other linspeeds at leftover indices of duplicate time_counts would be removed.
I'm not sure what else I can try. How can I adapt this (or find a new approach) to find all of the indices that correspond to duplicate values in an efficient manner? Any advice would be appreciated. (PS - I made numpy a tag because I think there is a way to do this via numpy that I haven't figured out yet.)
Without going into the details of how to implement and efficient rolling-window-maximum filter; reducing the duplicate values can be seen as a grouping-problem, which the numpy_indexed package (disclaimer: I am its author) provides efficient and simple solutions to:
import numpy_indexed as npi
unique_time, unique_speed = npi.group_by(time_count).max(linspeed)
For large input datasets (ie, where it matters), this should be a lot faster than any non-vectorized solution. Memory consumption is linear and performance in general NlogN; but since time_count appears to be sorted already, performance should be linear too.
OK, if you want to do this with numpy, best is to turn both of your lists into arrays:
l = np.array(linspeed)
tc = np.array(time_count)
Now, finding unique times is just an np.unique call:
u, i, c = np.unique(tc, return_inverse = True, return_counts = True)
u
Out[]: array([ 4., 6., 8., 10., 14., 16.])
i
Out[]: array([0, 1, 2, 2, 3, 3, 3, 4, 5], dtype=int32)
c
Out[]: array([1, 1, 2, 3, 1, 1])
Now you can either build your maximums with a for loop
m = np.array([np.max(l[i==j]) if c[j] > 1 else l[j] for j in range(u.size)])
m
Out[]: array([ 280. , 275. , 475.2, 400.9, 360.1, 400.9])
Or try some 2d method. This could be faster, but it would need to be optimized. This is just the basic idea.
np.max(np.where(i[None, :] == np.arange(u.size)[:, None], linspeed, 0),axis = 1)
Out[]: array([ 280. , 275. , 475.2, 400.9, 323.8, 289.7])
Now your m and u vectors are the same length and include the output you want.

How to create a container in pymc3

I am trying to build a model for the likelihood function of a particular outcome of a Langevin equation (Brownian particle in a harmonic potential):
Here is my model in pymc2 that seems to work:
https://github.com/hstrey/BayesianAnalysis/blob/master/Langevin%20simulation.ipynb
#define the model/function to be fitted.
def model(x):
t = pm.Uniform('t', 0.1, 20, value=2.0)
A = pm.Uniform('A', 0.1, 10, value=1.0)
#pm.deterministic(plot=False)
def S(t=t):
return 1-np.exp(-4*delta_t/t)
#pm.deterministic(plot=False)
def s(t=t):
return np.exp(-2*delta_t/t)
path = np.empty(N, dtype=object)
path[0]=pm.Normal('path_0',mu=0, tau=1/A, value=x[0], observed=True)
for i in range(1,N):
path[i] = pm.Normal('path_%i' % i,
mu=path[i-1]*s,
tau=1/A/S,
value=x[i],
observed=True)
return locals()
mcmc = pm.MCMC( model(x) )
mcmc.sample( 20000, 2000, 10 )
The basic idea is that each point depends on the previous point in the chain (Markov chain). Btw, x is an array of data, N is its length, delta_t is the time step =0.01. Any idea how to implement this in pymc3? I tried:
# define the model/function for diffusion in a harmonic potential
DHP_model = pm.Model()
with DHP_model:
t = pm.Uniform('t', 0.1, 20)
A = pm.Uniform('A', 0.1, 10)
S=1-pm.exp(-4*delta_t/t)
s=pm.exp(-2*delta_t/t)
path = np.empty(N, dtype=object)
path[0]=pm.Normal('path_0',mu=0, tau=1/A, observed=x[0])
for i in range(1,N):
path[i] = pm.Normal('path_%i' % i,
mu=path[i-1]*s,
tau=1/A/S,
observed=x[i])
Unfortunately the model crashes as soon as I try to run it. I tried some pymc3 examples (tutorial) on my machine and this is working.
Thanks in advance. I am really hoping that the new samplers in pymc3 will help me with this model. I am trying to apply Bayesian methods to single-molecule experiments.
Rather than creating many individual normally-distributed 1-D variables in a loop, you can make a custom distribution (by extending Continuous) that knows the formula for computing the log likelihood of your entire path. You can bootstrap this likelihood formula off of the Normal likelihood formula that pymc3 already knows. See the built-in AR1 class for an example.
Since your particle follows the Markov property, your likelihood looks like
import theano.tensor as T
def logp(path):
now = path[1:]
prev = path[:-1]
loglik_first = pm.Normal.dist(mu=0., tau=1./A).logp(path[0])
loglik_rest = T.sum(pm.Normal.dist(mu=prev*ss, tau=1./A/S).logp(now))
loglik_final = loglik_first + loglik_rest
return loglik_final
I'm guessing that you want to draw a value for ss at every time step, in which case you should make sure to specify ss = pm.exp(..., shape=len(x)-1), so that prev*ss in the block above gets interpreted as element-wise multiplication.
Then you can just specify your observations with
path = MyLangevin('path', ..., observed=x)
This should run much faster.
Since I did not see an answer to my question, let me answer it myself. I came up with the following solution:
# now lets model this data using pymc
# define the model/function for diffusion in a harmonic potential
DHP_model = pm.Model()
with DHP_model:
D = pm.Gamma('D',mu=mu_D,sd=sd_D)
A = pm.Gamma('A',mu=mu_A,sd=sd_A)
S=1.0-pm.exp(-2.0*delta_t*D/A)
ss=pm.exp(-delta_t*D/A)
path=pm.Normal('path_0',mu=0.0, tau=1/A, observed=x[0])
for i in range(1,N):
path = pm.Normal('path_%i' % i,
mu=path*ss,
tau=1.0/A/S,
observed=x[i])
start = pm.find_MAP()
print(start)
trace = pm.sample(100000, start=start)
unfortunately, this code takes at N=50 anywhere between 6hours to 2 days to compile. I am running on a pretty fast PC (24Gb RAM) running Ubuntu. I tried to using the GPU but that runs slightly slower. I suspect memory problems since it uses 99.8% of the memory when running. I tried the same calculation with Stan and it only takes 2min to run.

Replicate IDL 'smooth' in Python 2.7

I have been trying to work out how to replicate IDL's smooth function in Python and I just can't get anything like the same results. (Disclaimer: It is probably 10 years since I touched this kind of mathematical problem so it has been dumped to make way for information like where to find the cheapest local fuel). I am trying to code this:
smooth(b,w,/nan)
where b is a 2D float array containing NANs (zeros - missing data - have also been converted to NAN).
From the IDL documents, it appears smooth uses a boxcar, so from scipy.ndimage.filters I have tried:
bsmooth = uniform_filter(b, w)
I am aware that there are some fundamental differences here:
the default edge behaviour from IDL is "the end points are copied
from the original array to the result with no smoothing" whereas I
don't seem to have the option to do this with the uniform filter.
Treatment of the NaN elements. In IDL, the /nan keyword seems to
mean that where possible the NaN values will be filled by the result
of the other points in the window. If there are no valid points to
generate a result, by a MISSING keyword. I thought I could
approximate this behaviour following the smoothing using
scipy.interpolate's NearestNDInterpolator (thanks to the brilliant
explanation by Alex on here:
filling gaps on an image using numpy and scipy)
Here is my test array:
>>>b array([[ 0.97599638, 0.93114936, 0.87070072, 0.5379253 ],
[ 0.34873217, nan, 0.40985891, 0.22407863],
[ nan, nan, nan, 0.67532134],
[ nan, nan, 0.85441768, nan]])
My answers bore not the SLIGHTEST resemblance to IDL, whether I use the /nan keyword or not.
IDL> smooth(b,2,/nan)
0.97599638 0.93114936 0.87070072 0.53792530
0.34873217 0.70728749 0.60817236 0.22407863
NaN 0.53766960 0.54091913 0.67532134
NaN NaN 0.85441768 NaN
IDL> smooth(b,2)
0.97599638 0.93114936 0.87070072 0.53792530
0.34873217 -NaN -NaN 0.22407863
-NaN -NaN -NaN 0.67532134
-NaN -NaN 0.85441768 NaN
I confess I find the scipy documentation rather sparse on detail so I have no idea if I am really doing what I think I doing. The fact that the two python approaches which I believed would both smooth the image give different answers suggests that things are not what I understood them to be.
>>>uniform_filter(b, 2)
array([[ 0.97599638, 0.95357287, 0.90092504, 0.70431301],
[ 0.66236428, nan, nan, nan],
[ nan, nan, nan, nan],
[ nan, nan, nan, nan]])
I thought it was a bit odd it was so empty so I tried this with an array of 100 elements (still using a window of 2) and output the images. The results (first image is 'b' second is 'bsmooth') are not quite what I was hoping for:
Going back to the smaller array and following the examples in: http://scipy.github.io/old-wiki/pages/Cookbook/SignalSmooth which I thought would give the same output as uniform_filter, I tried:
>>> box = np.array([1,1,1,1])
>>> box = box.reshape(2,2)
>>> box
array([[1, 1],
[1, 1]])
>>> bsmooth = scipy.signal.convolve2d(b,box,mode='same')
>>> print bsmooth
[[ 0.97599638 1.90714574 1.80185008 1.40862602]
[ 1.32472855 nan nan 2.04256356]
[ nan nan nan nan]
[ nan nan nan nan]]
Obviously I have completely misunderstood the scipy functions, maybe even the IDL one. If anyone can help me to replicate the IDL smooth function as closely as possible, I would be extremely grateful. I am under considerable time pressure to get a solution for this that doesn't rely on IDL and I am tossing a coin to decide whether to code the function from scratch or develop a very contagious illness.
How can I perform the same smoothing in python?
First: Please use matplotlib.pyplot.imshow with interpolation="none" that's nicer to look at and maybe with greyscale.
So for your example: There is actually no convolution (filter) within scipy and numpy that treat's NaN as missing values (they propagate them within the convolution). At least I've found none so far and your boundary-treatement is also (to my knowledge) not implemented. But the boundary could be just replaced afterwards.
If you want to do convolution with NaN you can for example use astropy.convolution.convolve. There NaNs are interpolated using the kernel of your filter. But their convolution has some drawbacks as well: Border handling like you want isn't implemented there neither and your kernel must be of odd shape and the sum of your kernel must not be zero (or very close to it)
For example:
from astropy.convolution import convolve
import numpy as np
array = np.random.uniform(10,100, (4,4))
array[1,1] = np.nan
kernel = np.ones((3,3))
convolve(array, kernel)
as an example an initial array of
array([[ 97.19514587, 62.36979751, 93.54811286, 30.23567842],
[ 51.02184613, nan, 46.14769821, 60.08088041],
[ 20.86482452, 42.39661484, 36.96961278, 96.89180175],
[ 45.54453509, 76.61274347, 46.44485141, 25.40985372]])
will become:
array([[ 266.9009961 , 406.59680717, 348.69637399, 230.01236989],
[ 330.16243546, 506.82785931, 524.95440336, 363.87378443],
[ 292.75477064, 422.31693304, 487.26826319, 311.94469828],
[ 185.41871792, 268.83318211, 324.72547798, 205.71611967]])
if you want to "normalize" it, astropy offers the normalize_kernel parameter:
convolved = convolve(array, kernel, normalize_kernel=True)
array([[ 29.58753936, 42.09982189, 49.31793529, 33.00203873],
[ 49.87040638, 65.67695002, 66.10447436, 40.44026448],
[ 52.51126383, 63.03914444, 60.85474739, 35.88011742],
[ 39.40188443, 46.82350749, 40.1380926 , 22.46090152]])
If you want to replace the "edge" values with the ones from the original array just replace them:
convolved[0,:] = array[0,:]
convolved[-1,:] = array[-1,:]
convolved[:,0] = array[:,0]
convolved[:,-1] = array[:,-1]
So that's what the existing packages offer (as far as I know it). If you want to learn a bit of Cython or numba you can easily write your own convolutions that is not much slower (only a factor of 2-10) than the numpy/scipy ones but does EXACTLY what you want without messing around.
Since this is not something that is available in the python packages and because I saw the question asked several times during my research without satisfactory answers, here is how I solved the issue.
Provided is a test version of my function that I'm off to tidy up. I am sure there will be better ways to do the things I have done as I'm still fairly new to Python - please do recommend any appropriate changes.
Plots use autumn colourmap just because it allowed me to see the NaNs clearly.
My results:
IDL propagate
0.033369284 0.067915268 0.96602046 0.85623550
0.30435592 NaN NaN 100.00000
0.94065958 NaN NaN 0.90966976
0.018516513 0.044460904 0.051047217 NaN
python propagate
[[ 3.33692829e-02 6.79152655e-02 9.66020487e-01 8.56235492e-01]
[ 3.04355923e-01 nan nan 1.00000000e+02]
[ 9.40659566e-01 nan nan 9.09669768e-01]
[ 1.85165123e-02 4.44609040e-02 5.10472165e-02 nan]]
IDL replace
0.033369284 0.067915268 0.96602046 0.85623550
0.30435592 0.47452110 14.829881 100.00000
0.94065958 0.33833817 17.002417 0.90966976
0.018516513 0.044460904 0.051047217 NaN
python replace
[[ 3.33692829e-02 6.79152655e-02 9.66020487e-01 8.56235492e-01]
[ 3.04355923e-01 4.74521092e-01 1.48298812e+01 1.00000000e+02]
[ 9.40659566e-01 3.38338177e-01 1.70024175e+01 9.09669768e-01]
[ 1.85165123e-02 4.44609040e-02 5.10472165e-02 nan]]
My function:
#!/usr/bin/env python
# smooth.py
__version__ = 0.1
# Version 0.1 29 Feb 2016 ELH Test release
import numpy as np
import matplotlib.pyplot as mp
def Smooth(v1, w, nanopt):
# v1 is the input 2D numpy array.
# w is the width of the square window along one dimension
# nanopt can be replace or propagate
'''
v1 = np.array(
[[3.33692829e-02, 6.79152655e-02, 9.66020487e-01, 8.56235492e-01],
[3.04355923e-01, np.nan , 4.86013025e-01, 1.00000000e+02],
[9.40659566e-01, 5.23314093e-01, np.nan , 9.09669768e-01],
[1.85165123e-02, 4.44609040e-02, 5.10472165e-02, np.nan ]])
w = 2
'''
mp.imshow(v1, interpolation='None', cmap='autumn')
mp.show()
# make a copy of the array for the output:
vout=np.copy(v1)
# If w is even, add one
if w % 2 == 0:
w = w + 1
# get the size of each dim of the input:
r,c = v1.shape
# Assume that w, the width of the window is always square.
startrc = (w - 1)/2
stopr = r - ((w + 1)/2) + 1
stopc = c - ((w + 1)/2) + 1
# For all pixels within the border defined by the box size, calculate the average in the window.
# There are two options:
# Ignore NaNs and replace the value where possible.
# Propagate the NaNs
for col in range(startrc,stopc):
# Calculate the window start and stop columns
startwc = col - (w/2)
stopwc = col + (w/2) + 1
for row in range (startrc,stopr):
# Calculate the window start and stop rows
startwr = row - (w/2)
stopwr = row + (w/2) + 1
# Extract the window
window = v1[startwr:stopwr, startwc:stopwc]
if nanopt == 'replace':
# If we're replacing Nans, then select only the finite elements
window = window[np.isfinite(window)]
# Calculate the mean of the window
vout[row,col] = np.mean(window)
mp.imshow(vout, interpolation='None', cmap='autumn')
mp.show()
return vout

How to set offset with matplotlib

I'm trying to remove the offset that matplotlib automatically put on my graphs. For example, with the following code:
x=np.array([1., 2., 3.])
y=2.*x*1.e7
MyFig = plt.figure()
MyAx = MyFig.add_subplot(111)
MyAx.plot(x,y)
I obtain the following result (sorry, I cannot post image): the y-axis have the ticks 2, 2.5, 3, ..., 6, with a unique "x10^7" at the top of the y axis.
I would like to remove the "x10^7" from the top of the axis, and making it appearing with each tick (2x10^7, 2.5x10^7, etc...). If I understood well what I saw in other topics, I have to play with the use_Offset variable. So I tried the following thing:
MyFormatter = MyAx.axes.yaxis.get_major_formatter()
MyFormatter.useOffset(False)
MyAx.axes.yaxis.set_major_formatter(MyFormatter)
without any success (result unchanged).
Am I doing something wrong? How can I change this behaviour? Or have I to manually set the ticks ?
Thanks by advance for any help !
You can use the FuncFormatter from the ticker module to format the ticklabels as you please:
import matplotlib.pyplot as plt
import numpy as np
from matplotlib.ticker import FuncFormatter
x=np.array([1., 2., 3.])
y=2.*x*1.e7
MyFig = plt.figure()
MyAx = MyFig.add_subplot(111)
def sci_notation(x, pos):
return "${:.1f} \\times 10^{{6}}$".format(x / 1.e7)
MyFormatter = FuncFormatter(sci_notation)
MyAx.axes.yaxis.set_major_formatter(MyFormatter)
MyAx.plot(x,y)
plt.show()
On a side note; the "x10^7" value that appears at the top of your axis is not an offset, but a factor used in scientific notation. This behavior can be disabled by calling MyFormatter.use_scientific(False). Numbers will then be displayed as decimals.
An offset is a value you have to add (or subtract) to the tickvalues rather than multiply with, as the latter is a scale.
For reference, the line
MyFormatter.useOffset(False)
should be
MyFormatter.set_useOffset(False)
as the first one is a bool (can only have the values True or False), which means it can not be called as a method. The latter is the method used to enable/disable the offset.