In information theory, multivariate mutual information (MMI) could be synergy (negative) or redundancy (positive). To simulate this two cases, assuming three variables X, Y and Z, all of them takes 0 or 1 (binary variable). And we repeat sampling them 12 times.
Case 1:
X = [ 0 0 0 0 0 0 1 1 1 1 1 1 ]
Y = [ 0 0 0 0 1 1 0 0 1 1 1 1 ]
Z = [ 0 0 1 1 1 1 0 0 0 0 1 1 ]
In this case, we assume a mechanism among XYZ taht when both Y and Z are 0 or 1, X takes 0 or 1 respectively. When Y = 0, Z = 1, then X takes 0, and Y = 1, Z = 0, then X takes 1.
The mmi(X,Y,Z) = -0.1699 in this case, indicating a synergy effect among three variable.
Case 2:
X = [ 0 0 0 0 0 0 1 1 1 1 1 1 ]
Y = [ 0 0 0 0 0 1 0 1 1 1 1 1 ]
Z = [ 0 1 1 1 1 1 0 0 0 0 0 1 ]
the machanism in this case is same as above. The difference is there are more samples of XY takes different value and less samples of both XY are 0 or 1.
The mmi(X,Y,Z) = 0.0333, indicating a redundancy.
So far, can I say in these two cases, synergy and redundancy show the similar mechanism (or relationship) among three variables? But how do we understand redundancy and particularly synergy in realistic data?
Related
I have a dataset 'df' that looks something like this:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6
A 1 0 0 1 0 1
B 1 1 0 0 1 0
C 1 1 1 0 0 1
D 0 0 1 0 0 1
As you can see there are several rows of ones and zeros. Can anyone suggest me a code in python such that I am able to count the number of times '1' occurs continuously before the first occurrence of a 1, 0 and 0 in order. For example, for member A, the first double zero event occurs at seen_2 and seen_3, so the event will be 1. Similarly for the member B, the first double zero event occurs at seen_3 and seen_4 so there are two 1s that occur before this. The resultant table should have a new column 'event' something like this:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6 event
A 1 0 0 1 0 1 1
B 1 1 0 0 1 0 2
C 1 1 1 0 0 1 3
D 0 0 1 0 0 1 1
My approach:
df = df.set_index('MEMBER')
# count 1 on each rows since the last 0
s = (df.stack()
.groupby(['MEMBER', df.eq(0).cumsum(1).stack()])
.cumsum().unstack()
)
# mask of the zeros:
u = s.eq(0)
# look for the first 1 0 0
idx = (~u &
u.shift(-1, axis=1, fill_value=False) &
u.shift(-2, axis=1, fill_value=False) ).idxmax(1)
# look up
df['event'] = s.lookup(idx.index, idx)
Test data:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6
0 A 1 0 1 0 0 1
1 B 1 1 0 0 1 0
2 C 1 1 1 0 0 1
3 D 0 0 1 0 0 1
4 E 1 0 1 1 0 0
Output:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6 event
0 A 1 0 1 0 0 1 1
1 B 1 1 0 0 1 0 2
2 C 1 1 1 0 0 1 3
3 D 0 0 1 0 0 1 1
4 E 1 0 1 1 0 0 2
I tried different ways but it seems impossible for me to do it efficiently without looping through.
Input is an array y and a percentage x.
e.g. input is
y=np.random.binomial(1,1,[10,10])
x=0.5
output
[[0 0 0 0 1 1 1 1 0 1]
[1 0 1 0 0 1 0 1 0 1]
[1 0 1 1 1 1 0 0 0 1]
[0 1 0 1 1 0 1 0 1 1]
[0 1 1 0 0 1 1 1 0 0]
[0 0 1 1 1 0 1 1 0 1]
[0 1 0 0 0 0 1 0 1 1]
[0 0 0 1 1 1 1 1 0 0]
[0 1 1 1 1 0 0 1 0 0]
[1 0 1 0 1 0 0 0 0 0]]
Here's one based on masking -
def set_nonzeros_to_zeros(a, setz_ratio):
nz_mask = a!=0
nz_count = nz_mask.sum()
z_set_count = int(np.round(setz_ratio*nz_count))
idx = np.random.choice(nz_count,z_set_count,replace=False)
mask0 = np.ones(nz_count,dtype=bool)
mask0.flat[idx] = 0
nz_mask[nz_mask] = mask0
a[~nz_mask] = 0
return a
We are skipping the generation all the indices with np.argwhere/np.nonzero in favor of a masking based one to focus on performance.
Sample run -
In [154]: np.random.seed(0)
...: a = np.random.randint(0,3,(5000,5000))
# number of non-0s before using solution
In [155]: (a!=0).sum()
Out[155]: 16670017
In [156]: a_out = set_nonzeros_to_zeros(a, setz_ratio=0.2) #set 20% of non-0s to 0s
# number of non-0s after using solution
In [157]: (a_out!=0).sum()
Out[157]: 13336014
# Verify
In [158]: 16670017 - 0.2*16670017
Out[158]: 13336013.6
There are a few vectorized methods that might help you, depending on what you want to do:
# Flatten the 2D array and get the indices of the non-zero elements
c = y.flatten()
d = c.nonzero()[0]
# Shuffle the indices and set the first 100x % to zero
np.random.shuffle(d)
x = 0.5
c[d[:int(x*len(d))]] = 0
# reshape to the original 2D shape
y = c.reshape(y.shape)
No doubt there are some efficiency improvements to be made here.
I want the simplest verb that gives a list of all boolean lists of given length.
e.g.
f=. NB. Insert magic here
f 2
0 0
0 1
1 0
1 1
f 3
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
This functionality has been recently added to the stats/base addon.
load 'stats/base/combinatorial' NB. or just load 'stats'
permrep 2 NB. permutations of size 2 from 2 items with replacement
0 0
0 1
1 0
1 1
3 permrep 2 NB. permutations of size 3 from 2 items with replacement
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
permrep NB. display definition of permrep
$:~ :(# #: i.#^~)
Using the Qt IDE you can view the script defining permrep and friends by entering open 'stats/base/combinatorial' in the Term window. Alternatively you can view it on Github.
To define f as specified in your question, the following should suffice:
f=: permrep&2
f=: (# #: i.#^~)&2 NB. alternatively
f 3
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
The #: ("Antibase 2") vocab page has an example close to what I want. I don't really understand that primitive but the following code gives a list of base 2 digits of the numbers 0 to 2^n-1:
f=. #:#i.#(2^])
(Thanks to Dan for getting me to look up #:.)
l have K (let K here be 7) distincts matrices of dimension (50,50).
I would like to create a new matrix L by filling it in diagonal with the K matrices. Hence L is of dimension (50*K,50*K).
What l have tried ?
K1=np.random.random((50,50))
N,N=K1.shape
K=7
out=np.zeros((K,N,K,N),K1.dtype)
np.einsum('ijik->ijk', out)[...] = K1
L=out.reshape(K*N, K*N) # L is of dimension (50*7,50*7)=(350,350)
Its indeed creating a new matrix L by stacking K1 seven times within its diagonal. However, l would like to stack respectively K1,K2,K3,K5,K6,K7 rather than K1 seven times.
Inputs :
K1=np.random.random((50,50))
K2=np.random.random((50,50))
K3=np.random.random((50,50))
K4=np.random.random((50,50))
K5=np.random.random((50,50))
K6=np.random.random((50,50))
K7=np.random.random((50,50))
L=np.zeros((50*7,50*7))#
Expected outputs :
L[:50,:50]=K1
L[50:100,50:100]=K2
L[100:150,100:50]=K3
L[150:200,150:200]=K4
L[200:250,200:250]=K5
L[250:300,250:300]=K6
L[300:350,300:350]=K7
You could try scipy.linalg.block_diag. If you look at the source, this function basically just loops over the given blocks the way you have written as your output. It can be used like:
K1=np.random.random((50,50))
K2=np.random.random((50,50))
K3=np.random.random((50,50))
K4=np.random.random((50,50))
K5=np.random.random((50,50))
K6=np.random.random((50,50))
K7=np.random.random((50,50))
L=sp.linalg.block_diag(K1,K2,K3,K4,K5,K6,K7)
If you have your K as a ndarray of shape (7,50,50) you can unpack it directly like:
K=np.random.random((7,50,50))
L=sp.linalg.block_diag(*K)
If you don't want to import scipy, you can always just write a simple loop to do what you have written for the expected output.
Here is a way to do that with NumPy:
import numpy as np
def put_in_diagonals(a):
n, rows, cols = a.shape
b = np.zeros((n * rows, n * cols), dtype=a.dtype)
a2 = a.reshape(-1, cols)
ii, jj = np.indices(a2.shape)
jj += (ii // rows) * cols
b[ii, jj] = a2
return b
# Test
a = np.arange(24).reshape(4, 2, 3)
print(put_in_diagonals(a))
Output:
[[ 0 1 2 0 0 0 0 0 0 0 0 0]
[ 3 4 5 0 0 0 0 0 0 0 0 0]
[ 0 0 0 6 7 8 0 0 0 0 0 0]
[ 0 0 0 9 10 11 0 0 0 0 0 0]
[ 0 0 0 0 0 0 12 13 14 0 0 0]
[ 0 0 0 0 0 0 15 16 17 0 0 0]
[ 0 0 0 0 0 0 0 0 0 18 19 20]
[ 0 0 0 0 0 0 0 0 0 21 22 23]]
I have a bag-of-words representation of a corpus stored in an D by W sparse matrix word_freqs. Each row is a document and each column is a word. A given element word_freqs[d,w] represents the number of occurrences of word w in document d.
I'm trying to obtain another D by W matrix not_word_occs where, for each element of word_freqs:
If word_freqs[d,w] is zero, not_word_occs[d,w] should be one.
Otherwise, not_word_occs[d,w] should be zero.
Eventually, this matrix will need to be multiplied with other matrices which might be dense or sparse.
I've tried a number of methods, including:
not_word_occs = (word_freqs == 0).astype(int)
This words for toy examples, but results in a MemoryError for my actual data (which is approx. 18,000x16,000).
I've also tried np.logical_not():
word_occs = sklearn.preprocessing.binarize(word_freqs)
not_word_occs = np.logical_not(word_freqs).astype(int)
This seemed promising, but np.logical_not() does not work on sparse matrices, giving the following error:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all().
Any ideas or guidance would be appreciated.
(By the way, word_freqs is generated by sklearn's preprocessing.CountVectorizer(). If there's a solution that involves converting this to another kind of matrix, I'm certainly open to that.)
The complement of the nonzero positions of a sparse matrix is dense. So if you want to achieve your stated goals with standard numpy arrays you will require quite a bit of RAM. Here's a quick and totally unscientific hack to give you an idea, how many arrays of that sort your computer can handle:
>>> import numpy as np
>>> a = []
>>> for j in range(100):
... print(j)
... a.append(np.ones((16000, 18000), dtype=int))
My laptop chokes at j=1. So unless you have a really good computer even if you can get the complement (you can do
>>> compl = np.ones(S.shape,int)
>>> compl[S.nonzero()] = 0
) memory will be an issue.
One way out may be to not explicitly compute the complement let's call it C = B1 - A, where B1 is the same-shape matrix completely filled with ones and A the adjacency matrix of your original sparse matrix. For example the matrix product XC can be written as XB1 - XA so you have one multiplication with the sparse A and one with B1 which is actually cheap because it boils down to computing row sums. The point here is that you can compute that without computing C first.
A particularly simple example would be multiplication with a one-hot vector. Such a multiplication just selects a column (if multiplying from the right) or row (if multiplying from the left) of the other matrix. Meaning you just need to find that column or row of the sparse matrix and take the complement (for a single slice no problem) and if you do this for a one-hot matrix, as above you needn't compute the complement explicitly.
Make a small sparse matrix:
In [743]: freq = sparse.random(10,10,.1)
In [744]: freq
Out[744]:
<10x10 sparse matrix of type '<class 'numpy.float64'>'
with 10 stored elements in COOrdinate format>
the repr(freq) shows the shape, elements and format.
In [745]: freq==0
/usr/local/lib/python3.5/dist-packages/scipy/sparse/compressed.py:213: SparseEfficiencyWarning: Comparing a sparse matrix with 0 using == is inefficient, try using != instead.
", try using != instead.", SparseEfficiencyWarning)
Out[745]:
<10x10 sparse matrix of type '<class 'numpy.bool_'>'
with 90 stored elements in Compressed Sparse Row format>
If do your first action, I get a warning and new array with 90 (out of 100) nonzero terms. That not is no longer sparse.
In general numpy functions do not work when applied to sparse matrices. To work they have to delegate the task to sparse methods. But even if logical_not worked it wouldn't solve the memory issue.
Here is an example of using Pandas.SparseDataFrame:
In [42]: X = (sparse.rand(10, 10, .1) != 0).astype(np.int64)
In [43]: X = (sparse.rand(10, 10, .1) != 0).astype(np.int64)
In [44]: d1 = pd.SparseDataFrame(X.toarray(), default_fill_value=0, dtype=np.int64)
In [45]: d2 = pd.SparseDataFrame(np.ones((10,10)), default_fill_value=1, dtype=np.int64)
In [46]: d1.memory_usage()
Out[46]:
Index 80
0 16
1 0
2 8
3 16
4 0
5 0
6 16
7 16
8 8
9 0
dtype: int64
In [47]: d2.memory_usage()
Out[47]:
Index 80
0 0
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 0
dtype: int64
math:
In [48]: d2 - d1
Out[48]:
0 1 2 3 4 5 6 7 8 9
0 1 1 0 0 1 1 0 1 1 1
1 1 1 1 1 1 1 1 1 0 1
2 1 1 1 1 1 1 1 1 1 1
3 1 1 1 1 1 1 1 0 1 1
4 1 1 1 1 1 1 1 1 1 1
5 0 1 1 1 1 1 1 1 1 1
6 1 1 1 1 1 1 1 1 1 1
7 0 1 1 0 1 1 1 0 1 1
8 1 1 1 1 1 1 0 1 1 1
9 1 1 1 1 1 1 1 1 1 1
source sparse matrix:
In [49]: d1
Out[49]:
0 1 2 3 4 5 6 7 8 9
0 0 0 1 1 0 0 1 0 0 0
1 0 0 0 0 0 0 0 0 1 0
2 0 0 0 0 0 0 0 0 0 0
3 0 0 0 0 0 0 0 1 0 0
4 0 0 0 0 0 0 0 0 0 0
5 1 0 0 0 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0
7 1 0 0 1 0 0 0 1 0 0
8 0 0 0 0 0 0 1 0 0 0
9 0 0 0 0 0 0 0 0 0 0
memory usage:
In [50]: (d2 - d1).memory_usage()
Out[50]:
Index 80
0 16
1 0
2 8
3 16
4 0
5 0
6 16
7 16
8 8
9 0
dtype: int64
PS if you can't build the whole SparseDataFrame at once (because of memory constraints), you can use an approach similar to one used in this answer