I'm a beginner to Python and I'm having some trouble with this. I have to make a for loop out of this problem. Can anyone explain how I would go about this?
nextNValues (startValue, increment, numberOfValues)
This function creates a string of numberOfValues values, starting with startValue and
counting by increment. For example, nextNValues (5, 4, 3) would generate a string of
(not including the comments):
5 - the start value
9 - counting by 4, the increment
13 - stopping after 3 lines of output, the numberOfValues
You could use range(startValue,startValue+(increment*numberofValues),increment).
for i in range(numberOfValues):
print startValue + i * increment
I am not sure if that is exactly what you are looking for... but it is my suggestion based on the information you have posted.
It would probably be easiest to write a for loop with an index like i and use that to add i*increment ti start value and save the resulting value to a list. Have the loop run numberOfValues times. If this is homework it would be better for you to write out the actual code for yourself
Related
I will input numbers serially but not in consecutive cells, it would be random like this:
I want to highlight numbers from where the serial number is broken as shown (Since number 17 is missing so number 18 onwards all number are highlighted)
Please help me with a custom formula for the conditional formatting of this.
try this new approach:
=REGEXMATCH(A2&"", INDEX(TEXTJOIN("|", 1, "×",
SORT(IFERROR(1/(1/((IFERROR(SORT(UNIQUE(FLATTEN(IFERROR(SPLIT(A$2:A, ",")))))<>
SEQUENCE(MAX(FLATTEN(IFERROR(SPLIT(A$2:A, ","))))), TRUE))*
SEQUENCE(MAX(FLATTEN(IFERROR(SPLIT(A$2:A, ",")))))))), 1, 0))))
try:
=A2>INDEX(MAX(IFERROR((SEQUENCE(MAX(A$2:A))=SORT(A$2:A))*SEQUENCE(MAX(A$2:A)))))
update:
=REGEXEXTRACT(A2&"", "\d+,?")*1>INDEX(MAX(IFERROR((SEQUENCE(MAX(A$2:A))=
SORT(UNIQUE(FLATTEN(SPLIT(A$2:A, ",")))))*SEQUENCE(MAX(A$2:A)))))
I have used substring function in composer and the output should be the number extracted from the given value. Here the condition reaches to false branch.
By following the doc I have used below steps but I am not getting the output. It will be very helpful if somebody provides the solution.
I have used "= substring('${dialog.Str}', 4, 7) " as well.
I want 1234 only intger part to be stored in a property.
Thanks...
use like this ${substring(dialog.str, 4,4)} .. you will get the answers 1234 ..
substring - (string, starting Idx, total length)
I have a matrix consisting of 3 rows and 4 columns of which which I require the central two columns.
I have attempted extracting the central two columns as follows:
a = a[[2 ;; 3, All]];
On the mathematica function list, the first entry in a[[2 ;; 3, All]] represents the rows and the second the columns, however whenever I try a[[All,2 ;; 3]] it removes the top row rather than the two columns. For some reason they seem inverted. I tried going around this by switching the entries around however, when I use a[[2 ;; 3, All]], I get the error: Part: Cannot take positions 2 through 3 in a.
I cannot wrap my head around why this keeps happening. It also refuses to extract single columns from the matrix as well.
You show that you are assigning a variable to itself and then saying that things don't work for you. That makes me think you might have previously made assignments to variables and the results of that are lurking in the background and might be responsible for what you are seeing.
With a fresh start of Mathematica, before you do anything else, try
mat={{a,b,c,d},
{e,f,g,h},
{i,j,k,l}};
take23[row_]:=Take[row,{2,3}];
newmat = Map[take23, mat]
Map performs the function take23 on every row and returns a list containing all the results giving
{{b,c},
{f,g},
{j,k}}
If need be you can abbreviate that to
newmat = Map[Take[#,{2,3}]&, mat]
but that requires you understand # and & and it gives the same result.
If necessary you can further abbreviate that to
newmat = Take[#,{2,3}]& /# mat
Map is widely used in Mathematica programming and can do many more things than just extract elements. Learning how to use that will increase your Mathematica skill greatly.
Or if you really need to use ;; then this
newmat = mat[[All, 2;;3]]
I interpret the documentation for that to mean you want to do something with All the rows and then within each row you want to extract from the second to the third item. That seems to work for me and instantly returns the same result.
If you instead wrote
newmat = mat[[1;;2, 2;;3]]
that would tell it that you wanted to work from row 1 down to row 2 and within those you want to work from column 2 to column 3 and that gives
{{b,c},
{f,g}}
Basically, what I want to do is look at the range information of a unified diff and know exactly which lines of code I should pay attention to.
For instance, this:
## -1827,7 +1827,7 ##
This tells me that in total only 1 line has changed, because the diff shows 3 lines above and below the change (so 7 - 6 = 1), and it also points me to the line 1830 (i.e. 1827 + 3).
To be more pedantic, this particular range information actually tells me that at line 1830, a line was removed (-), and at line 1830 a line was added (+).
Or to make that more obvious consider this range information for another diff:
## -878,15 +878,13 ##
What this is telling me is that at line 881 (878 + 3) 9 lines were deleted (15 - 6), but at line 881 only 7 lines were added (13 - 6).
So the question is, using a regex or some other Ruby string method, how do I pull out the above information easily?
i.e. how do I easily pull out this info:
Both The line numbers (i.e. just the 1827 or 878), which I can then add + 3 to determine the actual inline number I care about. It has to be both because both lines may not always be identical.
The number of lines affected (aka the 7, 15 or 13 right after the , in the above examples)
While I do that, how do I make sure to track the operation (addition or deletion) for each of the operations.
I tried slicing the string and going directly for a character -- e.g. myString[3] which gives me -, but that's the only character it reliably works for because the line numbers can be 1, 10, 100, 1000, 10000, etc. So the only way is to just scan the string and then parse it.
Edit 1
To add some code to show what I have tried.
Assume I have the contents of a diff in a variable called #diff_lines:
#diff_lines.each do |diff_line|
if diff_line.start_with?("##")
del_line_num_start = diff_line.split(/## /).second.split.first.split(/-/).second.split(/,/).first.to_i + 3
num_deleted_lines = diff_line.split(/## /).second.split.first.split(/-/).second.split(/,/).second.to_i - 6
add_line_num_start = diff_line.split(/## /).second.split.second.split(/\+/).second.split(/,/).first.to_i + 3
num_added_lines = diff_line.split(/## /).second.split.second.split(/\+/).second.split(/,/).second.to_i - 6
As you can see, the above works....but it is quite horrendous to look at and is OBVIOUSLY not very DRY.
Ideally I would like to be able to achieve the same thing, but just cleaner.
The general idea is to write a regular expression that has capture groups in it ((...)) to pick apart that string into something useful. For example:
diff_line.match(/\A##\s+\-(\d+),(\d+)\s+\+(\d+),(\d+)\s+##/)
This yields a MatchData object on a successful match. You can then apply this to some variables like:
if (m = diff_line.match(...))
a_start, a_len, b_start, b_len = m[1..4].map(&:to_i)
end
Then you can do whatever computations you need to do with these numbers.
If you're ever having trouble visualizing what a regular expression does, try a tool like Rubular to better illustrate the internals.
As an exercise in python lambdas (just so I can learn how to use them more properly) I gave myself an assignment to sort some strings based on something other than their natural string order.
I scraped apache for version number strings and then came up with a lambda to sort them based on numbers I extracted with regexes. It works, but I think it can be better I just don't know how to improve it so it's more robust.
from lxml import html
import requests
import re
# Send GET request to page and parse it into a list of html links
jmeter_archive_url='https://archive.apache.org/dist/jmeter/binaries/'
jmeter_archive_get=requests.get(url=jmeter_archive_url)
page_tree=html.fromstring(jmeter_archive_get.text)
list_of_links=page_tree.xpath('//a[#href]/text()')
# Filter out all the non-md5s. There are a lot of links, and ultimately
# it's more data than needed for his exercise
jmeter_md5_list=list(filter(lambda x: x.endswith('.tgz.md5'), list_of_links))
# Here's where the 'magic' happens. We use two different regexes to rip the first
# and then the second number out of the string and turn them into integers. We
# then return them in the order we grabbed them, allowing us to tie break.
jmeter_md5_list.sort(key=lambda val: (int(re.search('(\d+)\.\d+', val).group(1)), int(re.search('\d+\.(\d+)', val).group(1))))
print(jmeter_md5_list)
This does have the desired effect, The output is:
['jakarta-jmeter-2.5.1.tgz.md5', 'apache-jmeter-2.6.tgz.md5', 'apache-jmeter-2.7.tgz.md5', 'apache-jmeter-2.8.tgz.md5', 'apache-jmeter-2.9.tgz.md5', 'apache-jmeter-2.10.tgz.md5', 'apache-jmeter-2.11.tgz.md5', 'apache-jmeter-2.12.tgz.md5', 'apache-jmeter-2.13.tgz.md5']
So we can see that the strings are sorted into an order that makes sense. Lowest version first and highest version last. Immediate problems that I see with my solution are two-fold.
First, we have to create two different regexes to get the numbers we want instead of just capturing groups 1 and 2. Mainly because I know there are no multiline lambdas, I don't know how to reuse a single regex object instead of creating a second.
Secondly, this only works as long as the version numbers are two numbers separated by a single period. The first element is 2.5.1, which is sorted into the correct place but the current method wouldn't know how to tie break for 2.5.2, or 2.5.3, or for any string with an arbitrary number of version points.
So it works, but there's got to be a better way to do it. How can I improve this?
This is not a full answer, but it will get you far along the road to one.
The return value of the key function can be a tuple, and tuples sort naturally. You want the output from the key function to be:
((2, 5, 1), 'jakarta-jmeter')
((2, 6), 'apache-jmeter')
etc.
Do note that this is a poor use case for a lambda regardless.
Originally, I came up with this:
jmeter_md5_list.sort(key=lambda val: list(map(int, re.compile('(\d+(?!$))').findall(val))))
However, based on Ignacio Vazquez-Abrams's answer, I made the following changes.
def sortable_key_from_string(value):
version_tuple = tuple(map(int, re.compile('(\d+(?!$))').findall(value)))
match = re.match('^(\D+)', value)
version_name = ''
if match:
version_name = match.group(1)
return (version_tuple, version_name)
and this:
jmeter_md5_list.sort(key = lambda val: sortable_key_from_string(val))