if we write 12wkd3, how to choose/filter 123 as integer in octave?
example in octave:
A = input("A?\n")
A?
12wkd3
A = 123
while 12wkd3 is user keyboard input and A = 123 is the expected answer.
assuming that the general form you're looking for is taking an arbitrary string from the user input, removing anything non-numeric, and storing the result it as an integer:
A = input("A? /n",'s');
A = int32(str2num(A(isdigit(A))));
example output:
A?
324bhtk.p89u34
A = 3248934
to clarify what's written above:
in the input statement, the 's' argument causes the answer to get stored as a string, otherwise it's evaluated by Octave first. most inputs would produce errors, others may be interpreted as functions or variables.
isdigit(A) produces a logical array of values for A with a 1 for any character that is a 0-9 number, and 0 otherwise.
isdigit('a1 3 b.') = [0 1 0 1 0 0 0]
A(isdigit(A)) will produce a substring from A using only those values corresponding to a 1 in the logical array above.
A(isdigit(A)) = 13
that still returns a string, so you need to convert it into a number using str2num(). that, however, outputs a double precision number. so finally to get it to be an integer you can use int32()
Related
I am working on a problem in Python and don't understand the answer.
for number in range(1, 10):
if number % 2 == 0:
print(number)
The answer to this problem is 2,4,6,8
Can anyone explain this answer?
range is a function in python which generates a sequence of integers, for example:
r=range(3)
returns a iterable object range(0,3) which generates sequence of integers from 0 to 3-1(2),inorder for you to see the elements in it , you can loop through it:
for i in r:
print(i)
#prints number from 0 to 3-1
Or, wrap it in a list:
list(range(3)) //returns [0,1,2]
range can take 3 params as input start,end and optionally step.The parameters start and end are basically lower and upper bounds to the sequence.In the above example since we have given only one integer range considers start as 0 and end as 3. This function range(start,end,[step]) generates integers in the following manner: start,start+1....end-1 considering the above example 0,0+1...3-1
if you give both the start and the end params to the range, the function generates integers from start upto but not including end, Example:
for i in range(3,8):print(i) #prints numbers from 3 to 8-1
if you give the third parameter which is the step(which is usually 1 by default), then range adds that number to the sequence :
list(range(3,8)) or list(range(3,8,1)) # will return [3,4,5,6,7],sequence generation will be like:3,3+1,(3+1)+1...
list(range(3,8,2)) #returns [3,5,7];3,3+2,(3+2)+2....
So , coming to your question now :
for number in range(1, 10): if number % 2 == 0: print(number)
In the above code you are basically telling python to loop over the sequence of integeres between 1 to 9 and print the numbers which are divisible by 2,which prints 2,4,6,8.
Hope this helped you :)
I have a dataframe which has a column called hexa which has hex values like this. They are of dtype object.
hexa
0 00802259AA8D6204
1 00802259AA7F4504
2 00802259AA8D5A04
I would like to remove the first and last bits and reverse the values bitwise as follows:
hexa-rev
0 628DAA592280
1 457FAA592280
2 5A8DAA592280
Please help
I'll show you the complete solution up here and then explain its parts below:
def reverse_bits(bits):
trimmed_bits = bits[2:-2]
list_of_bits = [i+j for i, j in zip(trimmed_bits[::2], trimmed_bits[1::2])]
reversed_bits = [list_of_bits[-i] for i in range(1,len(list_of_bits)+1)]
return ''.join(reversed_bits)
df['hexa-rev'] = df['hexa'].apply(lambda x: reverse_bits(x))
There are possibly a couple ways of doing it, but this way should solve your problem. The general strategy will be defining a function and then using the apply() method to apply it to all values in the column. It should look something like this:
df['hexa-rev'] = df['hexa'].apply(lambda x: reverse_bits(x))
Now we need to define the function we're going to apply to it. Breaking it down into its parts, we strip the first and last bit by indexing. Because of how negative indexes work, this will eliminate the first and last bit, regardless of the size. Your result is a list of characters that we will join together after processing.
def reverse_bits(bits):
trimmed_bits = bits[2:-2]
The second line iterates through the list of characters, matches the first and second character of each bit together, and then concatenates them into a single string representing the bit.
def reverse_bits(bits):
trimmed_bits = bits[2:-2]
list_of_bits = [i+j for i, j in zip(trimmed_bits[::2], trimmed_bits[1::2])]
The second to last line returns the list you just made in reverse order. Lastly, the function returns a single string of bits.
def reverse_bits(bits):
trimmed_bits = bits[2:-2]
list_of_bits = [i+j for i, j in zip(trimmed_bits[::2], trimmed_bits[1::2])]
reversed_bits = [list_of_bits[-i] for i in range(1,len(list_of_bits)+1)]
return ''.join(reversed_bits)
I explained it in reverse order, but you want to define this function that you want applied to your column, and then use the apply() function to make it happen.
I am trying to iterate over rows in a Pandas Dataframe using the itertuples()-method, which works quite fine for my case. Now i want to check if a specific value ('x') is in a specific tuple. I used the count() method for that, as i need to use the number of occurences of x later.
The weird part is, for some Tuples that works just fine (i.e. in my case (namedtuple[7].count('x')) + (namedtuple[8].count('x')) ), but for some (i.e. namedtuple[9].count('x')) i get an AttributeError: 'int' object has no attribute 'count'
Would appreciate your help very much!
Apparently, some columns of your DataFrame are of object type (actually a string)
and some of them are of int type (more generally - numbers).
To count occurrences of x in each row, you should:
Apply a function to each row which:
checks whether the type of the current element is str,
if it is, return count('x'),
if not, return 0 (don't attempt to look for x in a number).
So far this function returns a Series, with a number of x in each column
(separately), so to compute the total for the whole row, this Series should
be summed.
Example of working code:
Test DataFrame:
C1 C2 C3
0 axxv bxy 10
1 vx cy 20
2 vv vx 30
Code:
for ind, row in df.iterrows():
print(ind, row.apply(lambda it:
it.count('x') if type(it).__name__ == 'str' else 0).sum())
(in my opinion, iterrows is more convenient here).
The result is:
0 3
1 1
2 1
So as you can see, it is possible to count occurrences of x,
even when some columns are not strings.
Not understanding this:
Number returned from DataReader: 185549633.66000035
We have a requirement to maintain the number of decimal places per a User Choice.
For example: maintain 7 places.
We are using:
FormatNumber(dr.Item("Field"), 7, TriState.false, , TriState.True)
The result is: 185,549,633.6600000.
We would like to maintain the 3 (or 35) at the end.
When subtracting two numbers from the resulting query we are getting a delta but trying to show these two numbers out to 6,7,8 digits is not working thus indicating a false delta to the user.
Any advice would be appreciated.
Based on my testing, you must be working with Double values rather than Decimal. Not surprisingly, the solution to your problem can be found in the documentation.
For a start, you should not be using FormatNumber. We're not in VB6 anymore ToTo. To format a number in VB.NET, call ToString on that number. I tested this:
Dim dbl = 185549633.66000035R
Dim dec = 185549633.66000035D
Dim dblString = dbl.ToString("n7")
Dim decString = dec.ToString("n7")
Console.WriteLine(dblString)
Console.WriteLine(decString)
and I saw the behaviour you describe, i.e. the output was:
185,549,633.6600000
185,549,633.6600004
I read the documentation for the Double.ToString method (note that FormatNumber would be calling ToString internally) and this is what it says:
By default, the return value only contains 15 digits of precision although a maximum of 17 digits is maintained internally. If the value of this instance has greater than 15 digits, ToString returns PositiveInfinitySymbol or NegativeInfinitySymbol instead of the expected number. If you require more precision, specify format with the "G17" format specification, which always returns 17 digits of precision, or "R", which returns 15 digits if the number can be represented with that precision or 17 digits if the number can only be represented with maximum precision.
I then tested this:
Dim dbl = 185549633.66000035R
Dim dblString16 = dbl.ToString("G16")
Dim dblString17 = dbl.ToString("G17")
Console.WriteLine(dblString16)
Console.WriteLine(dblString17)
and the result was:
185549633.6600004
185549633.66000035
1 - 2 of 2
Above is my text. This is from paging of a web application. How do i extract the last number of the above text. SO i will get the count of list in that page and i can run a loop with respect to the number.
You can use substring
Let's consider your example. You have a String 1 - 2 of 2 (pagination probably)
Each of individual character is a specified index of a String
1 = 0
space = 1
- = 2
space = 3
etc.
String has a set of methods to perform various tasks. One of them is length() which gives you number of characters in your String
What you can do is to pass your length of String to substring.
Example:
myString.substring(0,1) will give you results of 1
myString.substring(0,myString.length()) wil give you results of 1 - 2 of 5
Additional info: myString.length() is an int type so you can perform math operations like + or -
myString.substring(0,myString.length()-1) will give you results of 1 - 2 of
I gave you the tools, now it's time for you to find the solutions.
You could just split the string using the spaces and then grab the last element of the split array. That should cover you even if the last number has more than one digit. Throw in a trim, just in case, to remove any leading/trailing white space.
String[] splitter = pageCount.trim().split(" ");
System.out.println(splitter[splitter.length - 1]);