All boolean possibilities of given length in J - j

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 #:.)

Related

Count number of non zero columns in a given set of columns of a data frame - pandas

I have a df as shown below
df:
Id Jan20 Feb20 Mar20 Apr20 May20 Jun20 Jul20 Aug20 Sep20 Oct20 Nov20 Dec20 Amount
1 20 0 0 12 1 3 1 0 0 2 2 0 100
2 0 0 2 1 0 2 0 0 1 0 0 0 500
3 1 2 1 2 3 1 1 2 2 3 1 1 300
From the above I would like to calculate Activeness value which is the number of non zero columns in the month columns as given below.
'Jan20', 'Feb20', 'Mar20', 'Apr20', 'May20', 'Jun20', 'Jul20',
'Aug20', 'Sep20', 'Oct20', 'Nov20', 'Dec20'
Expected Output:
Id Jan20 Feb20 Mar20 Apr20 May20 Jun20 Jul20 Aug20 Sep20 Oct20 Nov20 Dec20 Amount Activeness
1 20 0 0 12 1 3 1 0 0 2 2 0 100 7
2 0 0 2 1 0 2 0 0 1 0 0 0 500 4
3 1 2 1 2 3 1 1 2 2 3 1 1 300 12
I tried below code:
df['Activeness'] = pd.Series(index=df.index, data=np.count_nonzero(df[['Jan20', 'Feb20',
'Mar20', 'Apr20', 'May20', 'Jun20', 'Jul20',
'Aug20', 'Sep20', 'Oct20', 'Nov20', 'Dec20']], axis=1))
which is working well, but I would like to know is there any method that is faster than this.

You can try:
df['Activeness'] = df.filter(like = '20').ne(0, axis =1).sum(1)

Pattern identification and sequence detection

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

how to remove outermost logic?

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)).

Is there a way to make one integer increase when a second integer increases by a set amount in python 3?

I'm trying to create a simple game in python 3 and I'm trying to build in an EXP system, for example, every 50 experience points, your health (Which is already an integer) increases by one. Is there a command for this?
(I'm coding this on repl.it if that matters)

I've never shunned guessing. :)
Let me suppose that you are incrementing a variable called experience_points and that, once for every 50 times you increment that you want to increment a variable called health by one.
experience_points += 1
if experience_points % 50 == 0:
health +=1
This bit of code shows how this might work. Notice how health goes up one for every 50 times that 'experience_points` goes up one.
Welcome to the modulus operator!
>>> experience_points = 0
>>> health = 0
>>> while True:
... # do something in the game
... experience_points += 1
... if experience_points % 50 == 0:
... health += 1
... print (experience_points, health, '<--', end='')
... if experience_points > 160:
... break
...
1 0 <--2 0 <--3 0 <--4 0 <--5 0 <--6 0 <--7 0 <--8 0 <--9 0 <--10 0 <--11 0 <--12 0 <--13 0 <--14 0 <--15 0 <--16 0 <--17 0 <--18 0 <--19 0 <--20 0 <--21 0 <--22 0 <--23 0 <--24 0 <--25 0 <--26 0 <--27 0 <--28 0 <--29 0 <--30 0 <--31 0 <--32 0 <--33 0 <--34 0 <--35 0 <--36 0 <--37 0 <--38 0 <--39 0 <--40 0 <--41 0 <--42 0 <--43 0 <--44 0 <--45 0 <--46 0 <--47 0 <--48 0 <--49 0 <--50 1 <--51 1 <--52 1 <--53 1 <--54 1 <--55 1 <--56 1 <--57 1 <--58 1 <--59 1 <--60 1 <--61 1 <--62 1 <--63 1 <--64 1 <--65 1 <--66 1 <--67 1 <--68 1 <--69 1 <--70 1 <--71 1 <--72 1 <--73 1 <--74 1 <--75 1 <--76 1 <--77 1 <--78 1 <--79 1 <--80 1 <--81 1 <--82 1 <--83 1 <--84 1 <--85 1 <--86 1 <--87 1 <--88 1 <--89 1 <--90 1 <--91 1 <--92 1 <--93 1 <--94 1 <--95 1 <--96 1 <--97 1 <--98 1 <--99 1 <--100 2 <--101 2 <--102 2 <--103 2 <--104 2 <--105 2 <--106 2 <--107 2 <--108 2 <--109 2 <--110 2 <--111 2 <--112 2 <--113 2 <--114 2 <--115 2 <--116 2 <--117 2 <--118 2 <--119 2 <--120 2 <--121 2 <--122 2 <--123 2 <--124 2 <--125 2 <--126 2 <--127 2 <--128 2 <--129 2 <--130 2 <--131 2 <--132 2 <--133 2 <--134 2 <--135 2 <--136 2 <--137 2 <--138 2 <--139 2 <--140 2 <--141 2 <--142 2 <--143 2 <--144 2 <--145 2 <--146 2 <--147 2 <--148 2 <--149 2 <--150 3 <--151 3 <--152 3 <--153 3 <--154 3 <--155 3 <--156 3 <--157 3 <--158 3 <--159 3 <--160 3 <--161 3 <--

Matlab string operation

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

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