My goal is to compute a derivative of a moving window of a multidimensional dataset along a given dimension, where the dataset is stored as Xarray DataArray or DataSet.
In the simplest case, given a 2D array I would like to compute a moving difference across multiple entries in one dimension, e.g.:
data = np.kron(np.linspace(0,1,10), np.linspace(1,4,6) ).reshape(10,6)
T=3
reducedArray = np.zeros_like(data)
for i in range(data.shape[1]):
if i < T:
reducedArray[:,i] = data[:,i] - data[:,0]
else:
reducedArray[:,i] = data[:,i] - data[:,i-T]
where the if i <T condition ensures that input and output contain proper values (i.e., no nans) and are of identical shape.
Xarray's diff aims to perform a finite-difference approximation of a given derivative order using nearest-neighbours, so it is not suitable here, hence the question:
Is it possible to perform this operation using Xarray functions only?
The rolling weighted average example appears to be something similar, but still too distinct due to the usage of NumPy routines. I've been thinking that something along the lines of the following should work:
xr2DDataArray = xr.DataArray(
data,
dims=('x','y'),
coords={'x':np.linspace(0,1,10), 'y':np.linspace(1,4,6)}
)
r = xr2DDataArray.rolling(x=T,min_periods=2)
r.reduce( redFn )
I am struggling with the definition of redFn here ,though.
Caveat The actual dataset to which the operation is to be applied will have a size of ~10GiB, so a solution that does not blow up the memory requirements will be highly appreciated!
Update/Solution
Using Xarray rolling
After sleeping on it and a bit more fiddling the post linked above actually contains a solution. To obtain a finite difference we just have to define the weights to be $\pm 1$ at the ends and $0$ else:
def fdMovingWindow(data, **kwargs):
T = kwargs['T'];
del kwargs['T'];
weights = np.zeros(T)
weights[0] = -1
weights[-1] = 1
axis = kwargs['axis']
if data.shape[axis] == T:
return np.sum(data * weights, **kwargs)
else:
return 0
r.reduce(fdMovingWindow, T=4)
alternatively, using construct and a dot product:
weights = np.zeros(T)
weights[0] = -1
weights[-1] = 1
xrWeights = xr.DataArray(weights, dims=['window'])
xr2DDataArray.rolling(y=T,min_periods=1).construct('window').dot(xrWeights)
This carries a massive caveat: The procedure essentially creates a list arrays representing the moving window. This is fine for a modest 2D / 3D array, but for a 4D array that takes up ~10 GiB in memory this will lead to an OOM death!
Simplicistic - memory efficient
A less memory-intensive way is to copy the array and work in a way similar to NumPy's arrays:
xrDiffArray = xr2DDataArray.copy()
dy = xr2DDataArray.y.values[1] - xr2DDataArray.y.values[0] #equidistant sampling
for src in xr2DDataArray:
if src.y.values < xr2DDataArray.y.values[0] + T*dy:
xrDiffArray.loc[dict(y = src.y.values)] = src.values - xr2DDataArray.values[0]
else:
xrDiffArray.loc[dict(y = src.y.values)] = src.values - xr2DDataArray.sel(y = src.y.values - dy*T).values
This will produce the intended result without dimensional errors, but it requires a copy of the dataset.
I was hoping to utilise Xarray to prevent a copy and instead just chain operations that are then evaluated if and when values are actually requested.
A suggestion as to how to accomplish this will still be welcomed!
I have never used xarray, so maybe I am mistaken, but I think you can get the result you want avoiding using loops and conditionals. This is at least twice faster than your example for numpy arrays:
data = np.kron(np.linspace(0,1,10), np.linspace(1,4,6)).reshape(10,6)
reducedArray = np.empty_like(data)
reducedArray[:, T:] = data[:, T:] - data[:, :-T]
reducedArray[:, :T] = data[:, :T] - data[:, 0, np.newaxis]
I imagine the improvement will be higher when using DataArrays.
It does not use xarray functions but neither depends on numpy functions. I am confident that translating this to xarray will be straightforward, I know that it works if there are no coords, but once you include them, you get an error because of the coords mismatch (coords of data[:, T:] and of data[:, :-T] are different). Sadly, I can't do better now.
I am trying to replicate the data by multiplying every value with a range of values and save the results as CSV.
I have created a function "Replicate_Data" which takes the input numpy array and multiply with a random value between a range. What is the best way to create a 100 files and save as P3D1 , P4D1 and so on.
def Replicate_Data(data: np.ndarray) -> np.ndarray:
Rep_factor= random.uniform(-3,7)
data1 = data * Rep_factor
return data1
P2D1 = Replicate_Data(P1D1)
np.savetxt("P2D1.csv", P2D1, delimiter="," , dtype = complex)
Here is an example you can use as reference.
I generate toy data named toy, then I make n random values using np.random.uniform and call it randos, then I multiply these two objects to form out using numpy broadcasting. You could also do this multiplication in a loop (the same one you save in, in fact): depending on the size of your input array it could be very memory intensive as I've written it. A more complete answer probably depends on the shape of your input data.
import numpy as np
toy = np.random.random(size=(2,2)) # a toy input array
n = 100 # number of random values
randos = np.random.uniform(-3,7,size=n) # generate 100 uniform randoms
# now multiply all elements in toy by the randoms in randos
out = toy[None,...]*randos[...,None,None] # this depends on the shape.
# this will work only if toy has two dimensions. Otherwise requires modification
# it will take a lot of memory... 100*toy.nbytes worth
# now save in the loop..
for i,o in enumerate(out):
name = 'P{}D1'.format(str(i+1))
np.savetxt(name,o,delimiter=",")
# a second way without the broadcasting (slow, better on memory)
# more like 2*toy.nbytes
#for i,r in enumerate(randos):
# name = 'P{}D1'.format(str(i+1))
# np.savetxt(name,r*toy,delimiter=",")
I have a set of 46 years worth of rainfall data. It's in the form of 46 numpy arrays each with a shape of 145, 192, so each year is a different array of maximum rainfall data at each lat and lon coordinate in the given model.
I need to create a global map of tau values by doing an M-K test (Mann-Kendall) for each coordinate over the 46 years.
I'm still learning python, so I've been having trouble finding a way to go through all the data in a simple way that doesn't involve me making 27840 new arrays for each coordinate.
So far I've looked into how to use scipy.stats.kendalltau and using the definition from here: https://github.com/mps9506/Mann-Kendall-Trend
EDIT:
To clarify and add a little more detail, I need to perform a test on for each coordinate and not just each file individually. For example, for the first M-K test, I would want my x=46 and I would want y=data1[0,0],data2[0,0],data3[0,0]...data46[0,0]. Then to repeat this process for every single coordinate in each array. In total the M-K test would be done 27840 times and leave me with 27840 tau values that I can then plot on a global map.
EDIT 2:
I'm now running into a different problem. Going off of the suggested code, I have the following:
for i in range(145):
for j in range(192):
out[i,j] = mk_test(yrmax[:,i,j],alpha=0.05)
print out
I used numpy.stack to stack all 46 arrays into a single array (yrmax) with shape: (46L, 145L, 192L) I've tested it out and it calculates p and tau correctly if I change the code from out[i,j] to just out. However, doing this messes up the for loop so it only takes the results from the last coordinate in stead of all of them. And if I leave the code as it is above, I get the error: TypeError: list indices must be integers, not tuple
My first guess was that it has to do with mk_test and how the information is supposed to be returned in the definition. So I've tried altering the code from the link above to change how the data is returned, but I keep getting errors relating back to tuples. So now I'm not sure where it's going wrong and how to fix it.
EDIT 3:
One more clarification I thought I should add. I've already modified the definition in the link so it returns only the two number values I want for creating maps, p and z.
I don't think this is as big an ask as you may imagine. From your description it sounds like you don't actually want the scipy kendalltau, but the function in the repository you posted. Here is a little example I set up:
from time import time
import numpy as np
from mk_test import mk_test
data = np.array([np.random.rand(145, 192) for _ in range(46)])
mk_res = np.empty((145, 192), dtype=object)
start = time()
for i in range(145):
for j in range(192):
out[i, j] = mk_test(data[:, i, j], alpha=0.05)
print(f'Elapsed Time: {time() - start} s')
Elapsed Time: 35.21990394592285 s
My system is a MacBook Pro 2.7 GHz Intel Core I7 with 16 GB Ram so nothing special.
Each entry in the mk_res array (shape 145, 192) corresponds to one of your coordinate points and contains an entry like so:
array(['no trend', 'False', '0.894546014835', '0.132554125342'], dtype='<U14')
One thing that might be useful would be to modify the code in mk_test.py to return all numerical values. So instead of 'no trend'/'positive'/'negative' you could return 0/1/-1, and 1/0 for True/False and then you wouldn't have to worry about the whole object array type. I don't know what kind of analysis you might want to do downstream but I imagine that would preemptively circumvent any headaches.
Thanks to the answers provided and some work I was able to work out a solution that I'll provide here for anyone else that needs to use the Mann-Kendall test for data analysis.
The first thing I needed to do was flatten the original array I had into a 1D array. I know there is probably an easier way to go about doing this, but I ultimately used the following code based on code Grr suggested using.
`x = 46
out1 = np.empty(x)
out = np.empty((0))
for i in range(146):
for j in range(193):
out1 = yrmax[:,i,j]
out = np.append(out, out1, axis=0) `
Then I reshaped the resulting array (out) as follows:
out2 = np.reshape(out,(27840,46))
I did this so my data would be in a format compatible with scipy.stats.kendalltau 27840 is the total number of values I have at every coordinate that will be on my map (i.e. it's just 145*192) and the 46 is the number of years the data spans.
I then used the following loop I modified from Grr's code to find Kendall-tau and it's respective p-value at each latitude and longitude over the 46 year period.
`x = range(46)
y = np.zeros((0))
for j in range(27840):
b = sc.stats.kendalltau(x,out2[j,:])
y = np.append(y, b, axis=0)`
Finally, I reshaped the data one for time as shown:newdata = np.reshape(y,(145,192,2)) so the final array is in a suitable format to be used to create a global map of both tau and p-values.
Thanks everyone for the assistance!
Depending on your situation, it might just be easiest to make the arrays.
You won't really need them all in memory at once (not that it sounds like a terrible amount of data). Something like this only has to deal with one "copied out" coordinate trend at once:
SIZE = (145,192)
year_matrices = load_years() # list of one 145x192 arrays per year
result_matrix = numpy.zeros(SIZE)
for x in range(SIZE[0]):
for y in range(SIZE[1]):
coord_trend = map(lambda d: d[x][y], year_matrices)
result_matrix[x][y] = analyze_trend(coord_trend)
print result_matrix
Now, there are things like itertools.izip that could help you if you really want to avoid actually copying the data.
Here's a concrete example of how Python's "zip" might works with data like yours (although as if you'd used ndarray.flatten on each year):
year_arrays = [
['y0_coord0_val', 'y0_coord1_val', 'y0_coord2_val', 'y0_coord2_val'],
['y1_coord0_val', 'y1_coord1_val', 'y1_coord2_val', 'y1_coord2_val'],
['y2_coord0_val', 'y2_coord1_val', 'y2_coord2_val', 'y2_coord2_val'],
]
assert len(year_arrays) == 3
assert len(year_arrays[0]) == 4
coord_arrays = zip(*year_arrays) # i.e. `zip(year_arrays[0], year_arrays[1], year_arrays[2])`
# original data is essentially transposed
assert len(coord_arrays) == 4
assert len(coord_arrays[0]) == 3
assert coord_arrays[0] == ('y0_coord0_val', 'y1_coord0_val', 'y2_coord0_val', 'y3_coord0_val')
assert coord_arrays[1] == ('y0_coord1_val', 'y1_coord1_val', 'y2_coord1_val', 'y3_coord1_val')
assert coord_arrays[2] == ('y0_coord2_val', 'y1_coord2_val', 'y2_coord2_val', 'y3_coord2_val')
assert coord_arrays[3] == ('y0_coord2_val', 'y1_coord2_val', 'y2_coord2_val', 'y3_coord2_val')
flat_result = map(analyze_trend, coord_arrays)
The example above still copies the data (and all at once, rather than a coordinate at a time!) but hopefully shows what's going on.
Now, if you replace zip with itertools.izip and map with itertools.map then the copies needn't occur — itertools wraps the original arrays and keeps track of where it should be fetching values from internally.
There's a catch, though: to take advantage itertools you to access the data only sequentially (i.e. through iteration). In your case, it looks like the code at https://github.com/mps9506/Mann-Kendall-Trend/blob/master/mk_test.py might not be compatible with that. (I haven't reviewed the algorithm itself to see if it could be.)
Also please note that in the example I've glossed over the numpy ndarray stuff and just show flat coordinate arrays. It looks like numpy has some of it's own options for handling this instead of itertools, e.g. this answer says "Taking the transpose of an array does not make a copy". Your question was somewhat general, so I've tried to give some general tips as to ways one might deal with larger data in Python.
I ran into the same task and have managed to come up with a vectorized solution using numpy and scipy.
The formula are the same as in this page: https://vsp.pnnl.gov/help/Vsample/Design_Trend_Mann_Kendall.htm.
The trickiest part is to work out the adjustment for the tied values. I modified the code as in this answer to compute the number of tied values for each record, in a vectorized manner.
Below are the 2 functions:
import copy
import numpy as np
from scipy.stats import norm
def countTies(x):
'''Count number of ties in rows of a 2D matrix
Args:
x (ndarray): 2d matrix.
Returns:
result (ndarray): 2d matrix with same shape as <x>. In each
row, the number of ties are inserted at (not really) arbitary
locations.
The locations of tie numbers in are not important, since
they will be subsequently put into a formula of sum(t*(t-1)*(2t+5)).
Inspired by: https://stackoverflow.com/a/24892274/2005415.
'''
if np.ndim(x) != 2:
raise Exception("<x> should be 2D.")
m, n = x.shape
pad0 = np.zeros([m, 1]).astype('int')
x = copy.deepcopy(x)
x.sort(axis=1)
diff = np.diff(x, axis=1)
cated = np.concatenate([pad0, np.where(diff==0, 1, 0), pad0], axis=1)
absdiff = np.abs(np.diff(cated, axis=1))
rows, cols = np.where(absdiff==1)
rows = rows.reshape(-1, 2)[:, 0]
cols = cols.reshape(-1, 2)
counts = np.diff(cols, axis=1)+1
result = np.zeros(x.shape).astype('int')
result[rows, cols[:,1]] = counts.flatten()
return result
def MannKendallTrend2D(data, tails=2, axis=0, verbose=True):
'''Vectorized Mann-Kendall tests on 2D matrix rows/columns
Args:
data (ndarray): 2d array with shape (m, n).
Keyword Args:
tails (int): 1 for 1-tail, 2 for 2-tail test.
axis (int): 0: test trend in each column. 1: test trend in each
row.
Returns:
z (ndarray): If <axis> = 0, 1d array with length <n>, standard scores
corresponding to data in each row in <x>.
If <axis> = 1, 1d array with length <m>, standard scores
corresponding to data in each column in <x>.
p (ndarray): p-values corresponding to <z>.
'''
if np.ndim(data) != 2:
raise Exception("<data> should be 2D.")
# alway put records in rows and do M-K test on each row
if axis == 0:
data = data.T
m, n = data.shape
mask = np.triu(np.ones([n, n])).astype('int')
mask = np.repeat(mask[None,...], m, axis=0)
s = np.sign(data[:,None,:]-data[:,:,None]).astype('int')
s = (s * mask).sum(axis=(1,2))
#--------------------Count ties--------------------
counts = countTies(data)
tt = counts * (counts - 1) * (2*counts + 5)
tt = tt.sum(axis=1)
#-----------------Sample Gaussian-----------------
var = (n * (n-1) * (2*n+5) - tt) / 18.
eps = 1e-8 # avoid dividing 0
z = (s - np.sign(s)) / (np.sqrt(var) + eps)
p = norm.cdf(z)
p = np.where(p>0.5, 1-p, p)
if tails==2:
p=p*2
return z, p
I assume your data come in the layout of (time, latitude, longitude), and you are examining the temporal trend for each lat/lon cell.
To simulate this task, I synthesized a sample data array of shape (50, 145, 192). The 50 time points are taken from Example 5.9 of the book Wilks 2011, Statistical methods in the atmospheric sciences. And then I simply duplicated the same time series 27840 times to make it (50, 145, 192).
Below is the computation:
x = np.array([0.44,1.18,2.69,2.08,3.66,1.72,2.82,0.72,1.46,1.30,1.35,0.54,\
2.74,1.13,2.50,1.72,2.27,2.82,1.98,2.44,2.53,2.00,1.12,2.13,1.36,\
4.9,2.94,1.75,1.69,1.88,1.31,1.76,2.17,2.38,1.16,1.39,1.36,\
1.03,1.11,1.35,1.44,1.84,1.69,3.,1.36,6.37,4.55,0.52,0.87,1.51])
# create a big cube with shape: (T, Y, X)
arr = np.zeros([len(x), 145, 192])
for i in range(arr.shape[1]):
for j in range(arr.shape[2]):
arr[:, i, j] = x
print(arr.shape)
# re-arrange into tabular layout: (Y*X, T)
arr = np.transpose(arr, [1, 2, 0])
arr = arr.reshape(-1, len(x))
print(arr.shape)
import time
t1 = time.time()
z, p = MannKendallTrend2D(arr, tails=2, axis=1)
p = p.reshape(145, 192)
t2 = time.time()
print('time =', t2-t1)
The p-value for that sample time series is 0.63341565, which I have validated against the pymannkendall module result. Since arr contains merely duplicated copies of x, the resultant p is a 2d array of size (145, 192), with all 0.63341565.
And it took me only 1.28 seconds to compute that.
I am working on a kind of bootstrapping procedure for visual fixation data, and would be helped by the insights of others on this issue I am having. I suspect that either I'm missing something related to the functioning of the random number generator (random.randrange), or it shows my currently novice understanding of numpy array iteration and slicing. Being a psychologist with only hobby-level programming experience, i would not be surprised if it turns out I'm doing this in a really backwards way.
When you want to perform statistical analysis on visual fixation data, you often need to take center-bias into account, which is the bias whereby observers tend to fixate more to the center of an image at first and more randomly in the image later. This bias causes a temporal correlation between fixations, and an ROC-analysis (Receiver Operator Characteristic) performed on such data needs a baseline based on a specific kind of bootstrap method.
In this case, the data resides in a numpy array named original. This array is of shape (22, 800, 15, 2), where the dimensions indicate [observer, image, fixation (x, y)]. So, 15 fixations per observer per image.
In the bootstrap, we generally want to replace each fixation with another fixation that occurs somewhere in the set of all other images and all observers, but at the same time (in this case: the same fixation index, index 2 of original).
I think this means that we have to do the following:
create a new array of the same dimensions as original. This array will be called shuffled.
check if current x or y in original == NaN. If so, do not change this fixation. Otherwise continue;
choose a random fixation from the subset of original that satisfies the following index: [all observers, all images except the current image, current fixation]. Make sure it does not contain NaN, otherwise pick another random fixation until it does not contain NaN;
Set shuffled to the random fixation at the current location in original.
I have a function that takes array original and does what is described above with the slight modification that when only one of the original x, y pair is NaN, it only sets that x or y in the random fixation to np.nan. When I iterate through the loops I saw good results. After iterating through +- 10 loops I was satisfied as all data looked perfect, after which I proceeded to remove the raw_input() breakpoints I had set and let the function process all of the data without interruption. When I did so, I noticed that the function slows down each loop and grinds to a halt when it reaches observer=0 image=48.
My code is as follows:
for obs_index, obs in enumerate(original):
for img_index, img in enumerate(obs):
print obs_index, img_index
for fix_index, fix in enumerate(img):
# do the following because sometimes only x or y in the original is NaN
rand_fix = (np.nan, np.nan)
while np.isnan(rand_fix[0]) or np.isnan(rand_fix[1]):
rand_obs = randrange(observers)
rand_img = img_index
while rand_img == img_index:
rand_img = randrange(images)
rand_fix = original[rand_obs, rand_img, fix_index]
# do the following because sometimes only x or y in the original is NaN
if np.isnan(fix[0]):
rand_fix[0] = np.nan
if np.isnan(fix[1]):
rand_fix[1] = np.nan
shuffled[obs_index, img_index, fix_index] = rand_fix
When this function finishes, shuffled should contain correctly shuffled fixation data for use in ROC-analysis.
SOLVED
I came up with the following code, that no longer slows down:
for obs_index, obs in enumerate(original):
for img_index, img in enumerate(obs):
for fix_index, fix in enumerate(img):
x = fix[0]
y = fix[1]
rand_x = np.nan
rand_y = np.nan
if not(np.isnan(x) or np.isnan(y)):
while np.isnan(rand_x) or np.isnan(rand_y):
rand_obs = randrange(observers)
rand_img = img_index
while rand_img == img_index:
rand_img = randrange(images)
rand_x = original[rand_obs, rand_img, fix_index, 0]
rand_y = original[rand_obs, rand_img, fix_index, 1]
shuffled[obs_index, img_index, fix_index, 0] = rand_x
shuffled[obs_index, img_index, fix_index, 1] = rand_y
I also fixed the way the new fixation was assigned to the location in shuffled, to follow numpy indexing properly.