I want to list all numbers from one to another but I have no idea how. For example, I want the input to be something like 2 and -3 and for the output to be 2 1 0 -1 -2 -3.
for i in reversed(range(b, a+1)):
print(i)
i = i-1
Assuming you want the integer numbers, you can do something like this:
a = int(input('first number: '))
b = int(input('second number: '))
if a > b:
a, b = b, a
for i in range(a, b+1):
print(i)
Related
I'm totally new to kdb+/q, and I found this problem below quite confusing to me. Just to simplify, we say we have this one line function f returns an one-row table with preset values, and I want to run this function over a combination of inputs x and y, like dates (list) and metas (table, with columns like orderid, px, size etc).
Now, I listed two ways to do so below. Since the function f doesn't really use any of the input, I would suppose the order of x and y doesn't matter since the difference is just which one is passed to f before another and only when two inputs passed would f starts to operate.
But why I got error in the second way, i.e. table follows the list?
Any idea and explanation is much appreciated.
f: {[x;y]
([] m: enlist `M; n: enlist `N)
};
x: 1 2 3;
y: ([] a: 4 5 6; b: 7 8 9);
raze raze f ' [y] ' [x]; // this one works
raze raze f ' [x] ' [y]; // this one gives ERROR: length Explanation: Arguments do not conform
What you're doing is effectively equivalent to:
f:{y;1};
q)(f'[([]a:1 2 3;b:4 5 3)])#/:1 2 3
1 1 1
1 1 1
1 1 1
(using extra brackets to make it clear the order of operation).
In this situation each one reduces to
q)f'[([]a:1 2 3;b:4 5 3);1]
1 1 1
q)f'[([]a:1 2 3;b:4 5 3);2]
1 1 1
q)f'[([]a:1 2 3;b:4 5 3);3]
1 1 1
The "length" is ok here because the "y" values are atomic and kdb automatically expands those atomic values to match the length of the table. In order words, kdb treats these as:
q)f'[([]a:1 2 3;b:4 5 3);1 1 1]
1 1 1
q)f'[([]a:1 2 3;b:4 5 3);2 2 2]
1 1 1
q)f'[([]a:1 2 3;b:4 5 3);3 3 3]
1 1 1
However, when you change the order it becomes:
(f'[1 2 3])#/:([]a:1 2 3;b:4 5 3)
which is equivalent to:
f'[1 2 3;`a`b!1 4]
f'[1 2 3;`a`b!2 5]
f'[1 2 3;`a`b!3 3]
but now you do have a length problem because the dictionaries in the "y" variable are not atomic, they have length 2. Which doesn't match the length of the list (3).
You don’t say so but it looks like you are studying how to iterate a binary function f over list arguments, which has brought you to projecting f' onto x, which gives you a unary f'[x] that you then iterate over y. If that’s how we got here, what you want might be as simple as x f'y, which iterates f over corresponding items in x and y.
However, you mention combinations of inputs. If you want effectively a Cartesian product based on f, then combine the iterators Each Right and Each Left to get x f:/:\:y.
That returns a matrix. You have razed your result. Depending on your argument types, you might be able to use cross to generate all the argument pair combinations, and Apply Each .' to apply f to each pair:
f .' x cross y
I have a structure myS with several fields, including myField, which in turns includes several other fields such as BB. I need to count how many time *'R_value' appears in BB.
I have tried:
sum(myS.myField.BB = 'R_value')
and this:
count = 0;
for i = 1:numel(myS.myField)
number_of_element = numel(myS.myField(i).BB)=='R_value'
count = count+number_of_element;
end
but it doesn't work. Any suggestion?
If you are just checking if BB is that literal string, then your loop is just:
count = 0;
for i = 1:numel(myS.myField)
count = count+strcmp(myS.myField(i).BB,'R_value')
end
numel counts how many elements are. Zero is an element. so is False. Just sum the array.
count = 0;
for i = 1:numel(myS.myField)
number_of_element = sum(myS.myField(i).BB==R_value)
count = count+number_of_element;
end
Also note you had the parenthesis wrong, so you where counting how many BB where in total, then comparing that number to R_value. I am assuming R_value is a number.
e.g.:
myS.myField(1).BB=[1 2 3 4 1 1 1]
myS.myField(2).BB=[4 5 65 1]
R_value=1
I am trying to realize my idea in matlab.
I consider two column A and B.
A=data(:,1)
B=data(:,5)
the data look like:
A B
1 1
2 1
3 1
... ...
100 20
... ...
150 30
151 1
... ...
The values in column A are timepoints.
I start with the first element in column A. It schould be A(1,1) and look on the first element in the column B B(1,1). If B(1,1)==1its true,if not its false. Then I increase consider the second raw of the column A and second raw of the column B and so on until the last raw of A and B.
How can I construck this loop??
You can just consider B likes the following:
result = (B == 1);
The result would be the same size of B such as you want. Nowm you can get the value of A on result likes the following:
valid_times = A(result);
I need to combine the binary representation of 6 and 7 together:
bin1 = fliplr(de2bi(6));
bin2 = fliplr(de2bi(7));
bin1 =
1 1 0
bin2 =
1 1 1
after the combination the number should be
bin3 = 110111
Does anyone have any idea on how to do this?
As suggested you can just concatenate them
bin3 = [bin1, bin2]
However, if you really want them packed together without spaces you can do it like this:
bin3 = num2str([bin1, bin2]);
bin3 = bin3(bin3 ~= ' ')
If you want to turn them in to a number now you can use str2num()
I want to convert the decimal number 27 into binary such a way that , first the digit 2 is converted and its binary value is placed in an array and then the digit 7 is converted and its binary number is placed in that array. what should I do?
thanks in advance
That's called binary-coded decimal. It's easiest to work right-to-left. Take the value modulo 10 (% operator in C/C++/ObjC) and put it in the array. Then integer-divide the value by 10 (/ operator in C/C++/ObjC). Continue until your value is zero. Then reverse the array if you need most-significant digit first.
If I understand your question correctly, you want to go from 27 to an array that looks like {0010, 0111}.
If you understand how base systems work (specifically the decimal system), this should be simple.
First, you find the remainder of your number when divided by 10. Your number 27 in this case would result with 7.
Then you integer divide your number by 10 and store it back in that variable. Your number 27 would result in 2.
How many times do you do this?
You do this until you have no more digits.
How many digits can you have?
Well, if you think about the number 100, it has 3 digits because the number needs to remember that one 10^2 exists in the number. On the other hand, 99 does not.
The answer to the previous question is 1 + floor of Log base 10 of the input number.
Log of 100 is 2, plus 1 is 3, which equals number of digits.
Log of 99 is a little less than 2, but flooring it is 1, plus 1 is 2.
In java it is like this:
int input = 27;
int number = 0;
int numDigits = Math.floor(Log(10, input)) + 1;
int[] digitArray = new int [numDigits];
for (int i = 0; i < numDigits; i++) {
number = input % 10;
digitArray[numDigits - i - 1] = number;
input = input / 10;
}
return digitArray;
Java doesn't have a Log function that is portable for any base (it has it for base e), but it is trivial to make a function for it.
double Log( double base, double value ) {
return Math.log(value)/Math.log(base);
}
Good luck.