Consider an array a of positive integers 1 <= a[i] <= 10^9
We can perform an operation on this array which involves taking any integer x and subtracting or adding it to a subarray of a
The goal is to make the entire array 0
Some examples:
2 1 2 1 => 1 0 1 0 => 0 0 1 0 => 0 0 0 0
This requires 3 operations. First subtract 1 from all the elements (subarray=a[0..3]), then individually subtract 1 from single element subarrays (a[0..0], a[2..2])
Alternative way could've been:
2 1 2 1 => 2 1 1 0 => 2 0 0 0 => 0 0 0 0
=> 1 0 0 0 => 0 0 0 0
Example 2:
4 6 2 => 0 2 2 => 0 0 0
Example 3:
10 3 2 9 => 10 10 9 9 => 0 0 9 9 => 0 0 0 0
I searched a lot, not able to find the exact problem statement. The solution that I came across though (https://stackoverflow.com/a/68789827/4014182) talks about a divide and conquer solution wherein you realize that every time there is a 0 in the array, you can "split" the array. But how to get that 0 in the first place is beyond my understanding.
Related
I have data in excel which have two columns 'Peak Value' & 'Label'. I want to add value in 'Label' column based on 'Peak Value' column.
So, Input looks like below
Peak Value 0 0 0 88 0 0 88 0 0 88 0
Label 0 0 0 0 0 0 0 0 0 0 0
Input
Whenever the value in 'Peak Value' is greater than zero then it add 1 in 'Label' and replace all the zeros below it. For the next value greater than zero it should get incremented to 2 and replace all the zeros by 2.
So, the output will look like this:
Peak Value 0 0 0 88 0 0 88 0 0 88 0
Label 0 0 0 1 1 1 2 2 2 3 3
Output
and so on....
I tried writing function but I am only able to add 1 when the value is greater than 0 in 'Peak Value'.
def funct(row):
if row['Peak Value']>0:
val = 1
else:
val = 0
return val
df['Label']= df.apply(funct, axis=1)
May be you could try using cumsum and ffill:
import numpy as np
df['Labels'] = (df['Peak Value'] > 0).groupby(df['Peak Value']).cumsum()
df['Labels'] = df['Labels'].replace(0, np.nan).ffill().replace(np.nan, 0).astype(int)
Output:
Peak Value Labels
0 0 0
1 0 0
2 0 0
3 88 1
4 0 1
5 0 1
6 88 2
7 0 2
8 0 2
9 88 3
10 0 3
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 #:.)
how to remove outermost logic?
such as
input column D result
And(OR(A,B),C)
output column E binary number
OR(A,B)
A B C result(D)after extract(E)
0 0 0 0 0
0 0 1 0 0
0 1 0 0 1
0 1 1 1 1
1 0 0 0 1
1 0 1 1 1
1 1 0 0 1
1 1 1 1 1
i tried in excel
=IF(NOT(AND(D2,C2))=TRUE,1,0)
but can not remove outermost logic
result after extract
0 0 0 =IF(AND(OR(A2,B2),C2)=TRUE,1,0) =IF(OR(A2,B2)=TRUE,1,0) =IF(NOT(AND(D2,C2))=TRUE,1,0)
0 0 1 =IF(AND(OR(A3,B3),C3)=TRUE,1,0) =IF(OR(A3,B3)=TRUE,1,0) =IF(NOT(AND(D3,C3))=TRUE,1,0)
0 1 0 =IF(AND(OR(A4,B4),C4)=TRUE,1,0) =IF(OR(A4,B4)=TRUE,1,0) =IF(NOT(AND(D4,C4))=TRUE,1,0)
0 1 1 =IF(AND(OR(A5,B5),C5)=TRUE,1,0) =IF(OR(A5,B5)=TRUE,1,0) =IF(NOT(AND(D5,C5))=TRUE,1,0)
1 0 0 =IF(AND(OR(A6,B6),C6)=TRUE,1,0) =IF(OR(A6,B6)=TRUE,1,0) =IF(NOT(AND(D6,C6))=TRUE,1,0)
1 0 1 =IF(AND(OR(A7,B7),C7)=TRUE,1,0) =IF(OR(A7,B7)=TRUE,1,0) =IF(NOT(AND(D7,C7))=TRUE,1,0)
1 1 0 =IF(AND(OR(A8,B8),C8)=TRUE,1,0) =IF(OR(A8,B8)=TRUE,1,0) =IF(NOT(AND(D8,C8))=TRUE,1,0)
1 1 1 =IF(AND(OR(A9,B9),C9)=TRUE,1,0) =IF(OR(A9,B9)=TRUE,1,0) =IF(NOT(AND(D9,C9))=TRUE,1,0)
By "remove the outermost logic", I assume you want to remove the IF function.
One thing to note is that in a formula like =IF(AND(OR(A2,B2),C2)=TRUE,1,0) you never need the =TRUE test. =IF(AND(OR(A2,B2),C2),1,0) will work exactly the same.
There are a couple of ways to convert a boolean (i.e. true/false value) into an integer without the explicit IF. One is --AND(OR(A2,B2),C2). Another is int(AND(OR(A2,B2),C2)).
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
I have converted a string to binary as follows
message='hello my name is kamran';
messagebin=dec2bin(message);
Is there any method for storing it in array?
I am not really sure of what you want to do here, but if you need to concatenate the rows of the binary representation (which is a matrix of numchars times bits_per_char), this is the code:
message = 'hello my name is kamran';
messagebin = dec2bin(double(message));
linearmessagebin = reshape(messagebin',1,numel(messagebin));
Please note that the double conversion returns your ASCII code. I do not have access to a Matlab installation here, but for example octave complains about the code you provided in the original question.
NOTE
As it was kindly pointed out to me, you have to transpose the messagebin before "serializing" it, in order to have the correct result.
If you want the result as numeric matrix, try:
>> str = 'hello world';
>> b = dec2bin(double(str),8) - '0'
b =
0 1 1 0 1 0 0 0
0 1 1 0 0 1 0 1
0 1 1 0 1 1 0 0
0 1 1 0 1 1 0 0
0 1 1 0 1 1 1 1
0 0 1 0 0 0 0 0
0 1 1 1 0 1 1 1
0 1 1 0 1 1 1 1
0 1 1 1 0 0 1 0
0 1 1 0 1 1 0 0
0 1 1 0 0 1 0 0
Each row corresponds to a character. You can easily reshape it into to sequence of 0,1