So, I have a formula ( =INDEX(Sheet1.A1:F15,RANDBETWEEN(1,15),RANDBETWEEN(1,6)) ) that returns a random number in the sheet. But, how to run the formula until the returned number is less than or equal to 25 ?
I thought of using for..next.. but couldn't get it how to run ...
Welcome!
As #thebusybee pointed out in his comment, a macro for this task is much easier than using built-in functions. As rightly pointed out #tohuwawohu, pre-filtering the values makes things a lot easier. The macro code could be, for example, like this
Option Explicit
Function getRandValue(aValues As Variant, nTypeCriteria As Integer, dCriteriaValue As Variant) As Variant
Rem Params: aValues - array of values,
Rem nTypeCriteria - -2 less then, -1 not more, 0 equal, 1 not less, 2 more than
Rem dCriteriaValue - value to compare
Dim aTemp As Variant
Dim i As Long, j As Long, k As Long
Dim bGoodValue As Boolean
k = UBound(aValues,1)*UBound(aValues,2)
ReDim aTemp(1 To k)
k = 0
For i = 1 To UBound(aValues,1)
For j = 1 To UBound(aValues,2)
bGoodValue = False
Select Case nTypeCriteria
Case -2
bGoodValue = (aValues(i,j) < dCriteriaValue)
Case -1
bGoodValue = (aValues(i,j) <= dCriteriaValue)
Case 0
bGoodValue = (aValues(i,j) = dCriteriaValue)
Case 1
bGoodValue = (aValues(i,j) >= dCriteriaValue)
Case 2
bGoodValue = (aValues(i,j) > dCriteriaValue)
End Select
If bGoodValue Then
k = k+1
aTemp(k) = aValues(i,j)
EndIf
Next j
Next i
If k<1 Then
getRandValue = "No matching values"
ElseIf k=1 Then
getRandValue = aTemp(k)
Else
getRandValue = aTemp(Rnd()*(k-1)+1)
EndIf
End Function
Just put a call to this function in a cell in the form
=GETRANDVALUE(A1:F15;-1;25)
Related
So I have arr = randi([0,20],20,1). I want to show: If there are numbers less than 5, fprintf('Yes\n') only once. Im using a for loop (for i = 1 : length(arr)) and indexing it.
As your description, maybe you need if statement within for loop like below
for i = 1:length(arr)
if arr(i) < 5
fprintf('Yes\n');
break
end
end
If you want to print Yes once, you can try
if any(arr < 5)
fprintf('Yes\n')
endif
If you don't want to use break, the code below might be an option
for i = 1:min(find(arr <5))
if (arr(i) < 5)
fprintf('Yes\n');
end
end
You can use a break statement upon finding the first value under 5 and printing the Yes statement.
Using a break Statement:
arr = randi([0,20],20,1);
for i = 1: length(arr)
if arr(i) < 5
fprintf("Yes\n");
break;
end
end
Extension:
By Using any() Function:
Alternatively, if you'd like to concise it down without the need for a for-loop the any() function can be used to determine if any values within the array meet a condition in this case arr < 5.
arr = randi([0,20],20,1);
if(any(arr < 5))
fprintf("Yes\n");
end
By Using a While Loop:
Check = 0;
arr = randi([0,20],20,1);
i = 1;
while (Check == 0 && i < length(arr))
if arr(i) < 5
fprintf("Yes\n");
Check = 1;
end
i = i + 1;
end
I understand the whole code and
I just want to know why there has to be a -1 at the end of the range function.
I've been checking it out with pythontutor but I can't make it out.
#Given 2 strings, a and b, return the number of the positions where they
#contain the same length 2 substring. So "xxcaazz" and "xxbaaz" yields 3,
#since the "xx", "aa", and "az" substrings appear in the same place in
#both strings.
def string_match(a, b):
shorter = min(len(a), len(b))
count = 0
for i in range(shorter -1): #<<<<<<<<< This is -1 I don't understand.
a_sub = a[i:i+2]
b_sub = b[i:i+2]
if a_sub == b_sub:
count = count + 1
return count
string_match('xxcaazz', 'xxbaaz')
string_match('abc', 'abc')
string_match('abc', 'axc')
I expect to understand why there has to be a -1 at the end of the range function. I will appreciate your help and explanation!
The value indices of the for loop are counted since 0 so the final value actually would be the (size -1)
I have a question about removing duplicates in a table (rexx language), I am on netphantom applications that are using the rexx language.
I need a sample on how to remove the duplicates in a table.
I do have a thoughts on how to do it though, like using two loops for these two tables which are A and B, but I am not familiar with this.
My situation is:
rc = PanlistInsertData('A',0,SAMPLE)
TABLE A (this table having duplicate data)
123
1
1234
12
123
1234
I need to filter out those duplicates data into TABLE B like this:
123
1234
1
12
You can use lookup stem variables to test if you have already found a value.
This should work (note I have not tested so there could be syntax errors)
no=0;
yes=1
lookup. = no /* initialize the stem to no, not strictly needed */
j=0
do i = 1 to in.0
v = in.i
if lookup.v <> yes then do
j = j + 1
out.j = v
lookup.v = yes
end
end
out.0 = j
You can eliminate the duplicates by :
If InStem first element, Move the element to OutStem Else check all the OutStem elements for the current InStem element
If element is found, Iterate to the next InStem element Else add InStem element to OutStem
Code Snippet :
/*Input Stem - InStem.
Output Stem - OutStem.
Array Counters - I, J, K */
J = 1
DO I = 1 TO InStem.0
IF I = 1 THEN
OutStem.I = InStem.I
ELSE
DO K = 1 TO J
IF (InStem.I ?= OutStem.K) & (K = J) THEN
DO
J = J + 1
OutStem.J = InStem.I
END
ELSE
DO
IF (InStem.I == OutStem.K) THEN
ITERATE I
END
END
END
OutStem.0 = J
Hope this helps.
I am trying to compute and plot the distribution of bigrams frequencies
First I did generate all possible bigrams which gives 1296 bigrams
then i extract the bigrams from a given file and save them in words1
my question is how to compute the frequency of these 1296 bigrams for the file a.txt?
if there are some bigrams did not appear at all in the file, then their frequencies should be zero
a.txt is any text file
clear
clc
%************create bigrams 1296 ***************************************
chars ='1234567890abcdefghijklmonpqrstuvwxyz';
chars1 ='1234567890abcdefghijklmonpqrstuvwxyz';
bigram='';
for i=1:36
for j=1:36
bigram = sprintf('%s%s%s',bigram,chars(i),chars1(j));
end
end
temp1 = regexp(bigram, sprintf('\\w{1,%d}', 1), 'match');
temp2 = cellfun(#(x,y) [x '' y],temp1(1:end-1)', temp1(2:end)','un',0);
bigrams = temp2;
bigrams = unique(bigrams);
bigrams = rot90(bigrams);
bigram = char(bigrams(1:end));
all_bigrams_len = length(bigrams);
clear temp temp1 temp2 i j chars1 chars;
%****** 1. Cleaning Data ******************************
collection = fileread('e:\a.txt');
collection = regexprep(collection,'<.*?>','');
collection = lower(collection);
collection = regexprep(collection,'\W','');
collection = strtrim(regexprep(collection,'\s*',''));
%*******************************************************
temp = regexp(collection, sprintf('\\w{1,%d}', 1), 'match');
temp2 = cellfun(#(x,y) [x '' y],temp(1:end-1)', temp(2:end)','un',0);
words1 = rot90(temp2);
%*******************************************************
words1_len = length(words1);
vocab1 = unique(words1);
vocab_len1 = length(vocab1);
[vocab1,void1,index1] = unique(words1);
frequencies1 = hist(index1,vocab_len1);
I. Character counting problem for a string
bsxfun based solution for counting characters -
counts = sum(bsxfun(#eq,[string1-0]',65:90))
Output -
counts =
2 0 0 0 0 2 0 1 0 0 ....
If you would like to get a tabulate output of counts against each letter -
out = [cellstr(['A':'Z']') num2cell(counts)']
Output -
out =
'A' [2]
'B' [0]
'C' [0]
'D' [0]
'E' [0]
'F' [2]
'G' [0]
'H' [1]
'I' [0]
....
Please note that this was a case-sensitive counting for upper-case letters.
For a lower-case letter counting, use this edit to this earlier code -
counts = sum(bsxfun(#eq,[string1-0]',97:122))
For a case insensitive counting, use this -
counts = sum(bsxfun(#eq,[upper(string1)-0]',65:90))
II. Bigram counting case
Let us suppose that you have all the possible bigrams saved in a 1D cell array bigrams1 and the incoming bigrams from the file are saved into another cell array words1. Let us also assume certain values in them for demonstration -
bigrams1 = {
'ar';
'de';
'c3';
'd1';
'ry';
't1';
'p1'}
words1 = {
'de';
'c3';
'd1';
'r9';
'yy';
'de';
'ry';
'de';
'dd';
'd1'}
Now, you can get the counts of the bigrams from words1 that are present in bigrams1 with this code -
[~,~,ind] = unique(vertcat(bigrams1,words1));
bigrams_lb = ind(1:numel(bigrams1)); %// label bigrams1
words1_lb = ind(numel(bigrams1)+1:end); %// label words1
counts = sum(bsxfun(#eq,bigrams_lb,words1_lb'),2)
out = [bigrams1 num2cell(counts)]
The output on code run is -
out =
'ar' [0]
'de' [3]
'c3' [1]
'd1' [2]
'ry' [1]
't1' [0]
'p1' [0]
The result shows that - First element ar from the list of all possible bigrams has no find in words1 ; second element de has three occurrences in words1 and so on.
Hey similar to Dennis solution you can just use histc()
string1 = 'ASHRAFF'
histc(string1,'ABCDEFGHIJKLMNOPQRSTUVWXYZ')
this checks the number of entries in the bins defined by the string 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' which is hopefully the alphabet (just wrote it fast so no garantee). The result is:
Columns 1 through 21
2 0 0 0 0 2 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0
Columns 22 through 26
0 0 0 0 0
Just a little modification of my solution:
string1 = 'ASHRAFF'
alphabet1='A':'Z'; %%// as stated by Oleg Komarov
data=histc(string1,alphabet1);
results=cell(2,26);
for k=1:26
results{1,k}= alphabet1(k);
results{2,k}= data(k);
end
If you look at results now you can easily check rather it works or not :D
This answer creates all bigrams, loads in the file does a little cleanup, ans then uses a combination of unique and histc to count the rows
Generate all Bigrams
note the order here is important as unique will sort the array so this way it is created presorted so the output matches expectation;
[y,x] = ndgrid(['0':'9','a':'z']);
allBigrams = [x(:),y(:)];
Read The File
this removes capitalisation and just pulls out any 0-9 or a-z character then creates a column vector of these
fileText = lower(fileread('d:\loremipsum.txt'));
cleanText = regexp(fileText,'([a-z0-9])','tokens');
cleanText = cell2mat(vertcat(cleanText{:}));
create bigrams from file by shifting by one and concatenating
fileBigrams = [cleanText(1:end-1),cleanText(2:end)];
Get Counts
the set of all bigrams is added to our set (so the values are created for all possible). Then a value ∈{1,2,...,1296} is assigned to each unique row using unique's 3rd output. Counts are then created with histc with the bins equal to the set of values from unique's output, 1 is subtracted from each bin to remove the complete set bigrams we added
[~,~,c] = unique([fileBigrams;allBigrams],'rows');
counts = histc(c,1:1296)-1;
Display
to view counts against text
[allBigrams, counts+'0']
or for something potentially more useful...
[sortedCounts,sortInd] = sort(counts,'descend');
[allBigrams(sortInd,:), sortedCounts+'0']
ans =
or9
at8
re8
in7
ol7
te7
do6 ...
Did not look into the entire code fragment, but from the example at the top of your question, I think you are looking to make a histogram:
string1 = 'ASHRAFF'
nr = histc(string1,'A':'Z')
Will give you:
2 0 0 0 0 2 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0
(Got a working solution with hist, but as #The Minion shows histc is more easy to use here.)
Note that this solution only deals with upper case letters.
You may want to do something like so if you want to put lower case letters in their correct bin:
string1 = 'ASHRAFF'
nr = histc(upper(string1),'A':'Z')
Or if you want them to be shown separately:
string1 = 'ASHRaFf'
nr = histc(upper(string1),['a':'z' 'A':'Z'])
bi_freq1 = zeros(1,all_bigrams_len);
for k=1: vocab_len1
for i=1:all_bigrams_len
if char(vocab1(k)) == char(bigrams(i))
bi_freq1(i) = frequencies1(k);
end
end
end
I am trying to complete the following task:
Create
a
script
that
will
repeatedly
create
a
random
integer
K
in
the
range
of
0
to
20
until
every
case
has
been
entered
at
least
once.
You
have
3
possible
cases.
Case
A
is
entered
when
the
range
of
the
random
integer
K
is
between
or
equal
to
0
and
7.
Case
B
is
entered
when
the
range
of
the
random
integer
K
is
between
or
equal
to
8
and
14.
Case
C
is
entered
when
the
range
of
the
random
integer
K
is
between
or
equal
to
15
and
20.
Rules:
When
a
case
is
entered
you
must
print
to
the
user
“Congratulations
you
entered
Case
(A,
B,
or
C)”.
You
can
only
enter
each
case
once.
If
the
program
attempts
to
enter
the
same
case
more
than
once,
you
must
print.
“Invalid,
that
case
has
already
been
entered”.
The
program
will
end
once
all
the
cases
have
been
entered
and
the
program
will
print
“Good
job,
you
have
entered
all
cases”.
If
the
program
attempts
to
enter
any
already
entered
cases
more
than
3
times
(3
total
times
not
just
for
one
specific
case),
the
program
will
end
and
print
to
the
user
“That
random
generator
wasn’t
random
enough”.
Here is the code I have so fa. It has taken me a couple hours to debug. Am I approaching this the wrong way????Please let me know.
K = round(rand*(20))
flag = 0;
counterA =0;
counterB=0;
counterC=0;
switch K
case {0,1,2,3,4,5,6,7}
fprintf('Congratulations you entered Case A\n')
flag = 1;
counterA = 1
case {8,9,10,11,12,13,14}
fprintf('Congratulations you entered Case B\n')
flag =2;
counterB = 1
case {15,16,17,18,19,20}
fprintf ('Congratulations you entered Case C\n')
flag = 3;
counterC = 1
end
while flag == 1 || flag == 2 || flag ==3
K = round(rand*(20))
if K >=0 && K<=7 && flag==1
disp ('Invalid, that case has already been entered')
counterA = counterA+1
elseif K >=8 && K<=14 && flag ==2
disp ('Invalid, that case has already been entered')
counterB=counterB+1
elseif K >=15 && K<=20 && flag==3
disp ('Invalid, that case has already been entered')
counterC =counterC+1
elseif K >=0 && K<=7 && flag ~=1
counterA =counterA+1
flag == 1;
if counterA==1&&counterB~=2 ||counterA==1&&counterC~=2
fprintf('COngrats guacamole A\n')
end
elseif K >=8 && K<=14 && flag ~=2
counterB=counterB+1
flag == 2;
if counterB ==1&&counterA~=2||counterB==1&&counterC~=2
fprintf('COngratsavacado B\n')
end
elseif K >=15 && K<=20 && flag~=3
counterC=counterC+1
flag == 3;
if counterC==1&&counterA~=2||counterC==1&&counterB~=2
fprintf ('Congratscilantro C\n')
end
end
if counterA==1 && counterB==1 && counterC==1
flag=100;
disp('DONE')
elseif counterA == 3|| counterB==3 || counterC==3
disp ('That random generator wasnt random enough')
flag =99;
elseif counterA==2||counterB==2||counterC==2
disp('Inval')
end
Some words about your code:
Don't use variable names like counterA,counterB,counterC, use a array with 3 elements instead. In this case: You need only a total limit, thus one variable is enough.
rand*20 generates random values between 0 and 20, but using round(rand*20) causes a lower probability for 0 and 20. Use randi if you need integers.
Use "Start indent" to format your code clean, it makes it easier to read.
This is not a full solution, the part with the 3 errors is missing. I think you will get this on your own.
caseNames={'A','B','C'};
caseEntered=[false,false,false];
%while there exist a case which is not entered and limit is not reached, continue
while ~all(caseEntered)
K = randi([0,20]);
switch K
case {0,1,2,3,4,5,6,7}
cs=1;
case {8,9,10,11,12,13,14}
cs=2;
case {15,16,17,18,19,20}
cs=3;
end
if caseEntered(cs)
%case has previously been entered, tdb
else
%case is entered frist time
fprintf('Congratulations you entered Case %s\n',caseNames{cs});
caseEntered(cs)=true;
end
end