how to remove outermost logic? - excel

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)).

Related

Returning column header corresponding to matched value

need some help here.. I am looking to retrieve Gender from Sheet 2 corresponding to the name in Sheet 1.
Step 1 - Match the name in sheet 1 to sheet 2 (not all names in sheet 1 will be in sheet 2, mark NA for non matching names)
Step 2 - Look for the corresponding gender in sheet 2.
Step 3 - Retrieve the column header or the last number in the column header (1,2,3...6)
Sheet 1
Name
Gender
w
???
e
r
t
y
u
i
q
w
e
r
Sheet 2
Name
Male 1
Female 2
other 3
other 4
other 5
Do not know 6
w
1
0
0
0
0
0
a
0
0
0
0
0
1
q
1
0
0
0
0
0
r
0
1
0
0
0
0
e
1
0
0
0
0
0
t
0
0
0
0
1
0
y
0
0
0
0
0
1
u
0
1
0
0
0
0

with Office 365 we can use FILTER:
=IFERROR(FILTER($F$1:$K$1,INDEX($F$2:$K$9,MATCH(A2,$E$2:$E$9,0),0)=1),"No Match")
With older versions we can use another INDEX/MATCH:
=IFERROR(INDEX($F$1:$K$1,MATCH(1,INDEX($F$2:$K$9,MATCH(A2,$E$2:$E$9,0),0),0)),"No Match")

How to identify where a particular sequence in a row occurs for the first time

I have a dataframe in pandas, an example of which is provided below:
Person appear_1 appear_2 appear_3 appear_4 appear_5 appear_6
A 1 0 0 1 0 0
B 1 1 0 0 1 0
C 1 0 1 1 0 0
D 0 0 1 0 0 1
E 1 1 1 1 1 1
As you can see 1 and 0 occurs randomly in different columns. It would be helpful, if anyone can suggest me a code in python such that I am able to find the column number where the 1 0 0 pattern occurs for the first time. For example, for member A, the first 1 0 0 pattern occurs at appear_1. so the first occurrence will be 1. Similarly for the member B, the first 1 0 0 pattern occurs at appear_2, so the first occurrence will be at column 2. The resulting table should have a new column named 'first_occurrence'. If there is no such 1 0 0 pattern occurs (like in row E) then the value in first occurrence column will the sum of number of 1 in that row. The resulting table should look something like this:
Person appear_1 appear_2 appear_3 appear_4 appear_5 appear_6 first_occurrence
A 1 0 0 1 0 0 1
B 1 1 0 0 1 0 2
C 1 0 1 1 0 0 4
D 0 0 1 0 0 1 3
E 1 1 1 1 1 1 6
Thank you in advance.

I try not to reinvent the wheel, so I develop on my answer to previous question. From that answer, you need to use additional idxmax, np.where, and get_indexer
cols = ['appear_1', 'appear_2', 'appear_3', 'appear_4', 'appear_5', 'appear_6']
df1 = df[cols]
m = df1[df1.eq(1)].ffill(1).notna()
df2 = df1[m].bfill(1).eq(0)
m2 = df2 & df2.shift(-1, axis=1, fill_value=True)
df['first_occurrence'] = np.where(m2.any(1), df1.columns.get_indexer(m2.idxmax(1)),
df1.shape[1])
Out[540]:
Person appear_1 appear_2 appear_3 appear_4 appear_5 appear_6 first_occurrence
0 A 1 0 0 1 0 0 1
1 B 1 1 0 0 1 0 2
2 C 1 0 1 1 0 0 4
3 D 0 0 1 0 0 1 3
4 E 1 1 1 1 1 1 6

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

All boolean possibilities of given length in 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 #:.)

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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