I just found a CodeGolf answer here http://repl.it/2Om/6:
puts"The eight#{e="een-hundreds were a time for "}rum.
The ninet#{e}fun.
The two-thousands are a time to run
a civilized classroom.
"*(?X-??)
Original post: https://codegolf.stackexchange.com/a/40250
I am curious, how does it work? I have never seen (?X-??) before. What's happening here?
?Char gives the ASCII code of Char.
?X = "X".ord = 88
?? = "?".ord = 63
?X - ?? = 88-63 = 25
There is your integer: 25
Then "a"*25 = "aaaaaaaaaaaaaaaaaaaaaaaaa"
Related
I am studying for this great Coursera course https://www.coursera.org/learn/algorithmic-toolbox . On the fourth week, we have an assignment related to binary trees.
I think I did a good job. I created a binary search code that solves this problem using recursion in Python3. That's my code:
#python3
data_in_sequence = list(map(int,(input().split())))
data_in_keys = list(map(int,(input()).split()))
original_array = data_in_sequence[1:]
data_in_sequence = data_in_sequence[1:]
data_in_keys = data_in_keys[1:]
def binary_search(data_in_sequence,target):
answer = 0
sub_array = data_in_sequence
#print("sub_array",sub_array)
if not sub_array:
# print("sub_array",sub_array)
answer = -1
return answer
#print("target",target)
mid_point_index = (len(sub_array)//2)
#print("mid_point", sub_array[mid_point_index])
beg_point_index = 0
#print("beg_point_index",beg_point_index)
end_point_index = len(sub_array)-1
#print("end_point_index",end_point_index)
if sub_array[mid_point_index]==target:
#print ("final midpoint, ", sub_array[mid_point_index])
#print ("original_array",original_array)
#print("sub_array[mid_point_index]",sub_array[mid_point_index])
#print ("answer",answer)
answer = original_array.index(sub_array[mid_point_index])
return answer
elif target>sub_array[mid_point_index]:
#print("target num higher than current midpoint")
beg_point_index = mid_point_index+1
sub_array=sub_array[beg_point_index:]
end_point_index = len(sub_array)-1
#print("sub_array",sub_array)
return binary_search(sub_array,target)
elif target<sub_array[mid_point_index]:
#print("target num smaller than current midpoint")
sub_array = sub_array[:mid_point_index]
return binary_search(sub_array,target)
else:
return None
def bin_search_over_seq(data_in_sequence,data_in_keys):
final_output = ""
for key in data_in_keys:
final_output = final_output + " " + str(binary_search(data_in_sequence,key))
return final_output
print (bin_search_over_seq(data_in_sequence,data_in_keys))
I usually get the correct output. For instance, if I input:
5 1 5 8 12 13
5 8 1 23 1 11
I get the correct indexes of the sequences or (-1) if the term is not in sequence (first line):
2 0 -1 0 -1
However, my code does not pass on the expected running time.
Failed case #4/22: time limit exceeded (Time used: 13.47/10.00, memory used: 36696064/536870912.)
I think this happens not due to the implementation of my binary search (I think it is right). Actually, I think this happens due to some inneficieny in a peripheral part of the code. Like the way I am managing to output the final answer. However, the way I am presenting the final answer does not seem to be really "heavy"... I am lost.
Am I not seeing something? Is there another inefficiency I am not seeing? How can I solve this? Just trying to present the final result in a faster way?
I have a ruby string array value and i want to get it as string value. I am using ruby with chef recipe. Running in windows platform. Code-
version_string = Mixlib::ShellOut.new('some.exe -version').run_command
Log.info(version.stdout.to_s)
extract_var = version_string.stdout.to_s.lines.grep(/ver/)
Log.info('version:'+ extract_var.to_s)
output is coming-
version 530
[2016-06-08T07:03:49+00:00] INFO: version ["version 530\r\n"]
I want to extract 530 string only.
long time no see since Rot :)
You can use some Chef helper methods and regular expressions to make this a little easier.
output = shell_out!('saphostexec.exe -version', cwd: 'C:\\Program Files\\hostctrl\\exe').stdout
if output =~ /kernel release\s+(\d+)/
kernel_version = $1
else
raise "unable to parse kernel version"
end
Chef::Log.info(kernel_version)
As you want val = 720 and not val = "720" you can write
val = strvar.first.to_i
#=> 720
You can return the first series of digits found as an integer from the current_kernel string with String#[regexp] :
current_kernel[/\d+/].to_i
#=> 720
Hello stackexchange community.
I've built a simple tables converter, the main function of which is to convert the table from
1a Value
1b Value
1c Value
1d Value
to
a b c d
1 Value Value Value Value
Unfortunately, the macro runs pretty slow (~ 3 lines per second for one column).
I'd really appreciate if someone could take a look at my piece of code and suggest the way to speed it up.
Here's the piece of code:
Dim LastFinalList As Integer: LastFinalList = Sheet1.Range("O1000").End(xlUp).Row
For Col = 16 To 19
For c = 2 To LastFinalList
searchrange = Sheet1.Range("J:L")
lookfor = Sheet1.Cells(c, 15) & Sheet1.Cells(1, Col)
CountFor = Application.VLookup(lookfor, searchrange, 3, False)
If IsError(CountFor) Then
Sheet1.Cells(c, Col).Value = "0"
Else
Sheet1.Cells(c, Col).Value = CountFor
End If
Next c
Next Col
Thanks in advance and best regards!
UPD:
The Data in unconverted table looks like this (e.g):
Updated by Macro
Value Number Type Key Count Average Value
10 1 a 1a 2 20
30 1 a 1a 2 20
40 1 b 1b 1 40
50 1 c 1c 1 50
So it is also required to calculate averages of repeating types, create a unique list of Numbers (which is LastFinalList in my case) and finally convert it to this:
Number a b c
1 20 40 50
application.vlookupseraches by Number&Type Key, which is also assigned in the unconverted table by macro. The same time those Keys are counted, in order to calculate average for the repeating ones.
Everything works in a blink of an eye till it comes to 'to update final table part.
Full Code:
Sub ConvertToTable()
Dim LastMeter As Integer: LastMeter = Sheet1.Range("I1000").End(xlUp).Row
Sheet1.Range(Cells(2, 9), Cells(LastMeter, 9)).AdvancedFilter Action:=xlFilterCopy, CopyToRange:=Sheet1.Range("O2"), Unique:=True
Sheet1.Range("O1").Value = "The List"
Sheet1.Range("O2").Delete Shift:=xlUp
' to assign keys
For i = 2 To LastMeter
Set CountOpt = Sheet1.Cells(i, 10)
Sheet1.Cells(i, 10).FormulaR1C1 = "=r[0]c[-1]&r[0]c[-2]"
Sheet1.Cells(i, 11).FormulaR1C1 = "=COUNTIF(c10:c10, r[0]c10)"
Next i
'to calculate averages
For x = 2 To LastMeter
If Sheet1.Cells(x, 11).Value = 1 Then
Sheet1.Cells(x, 12).FormulaR1C1 = "=rc7"
ElseIf Sheet1.Cells(x, 11).Value > 1 Then
If Sheet1.Cells(x, 10).Value <> Sheet1.Cells(x - 1, 10).Value Then
Sheet1.Cells(x, 12).FormulaR1C1 = "=ROUND((SUM(rc7:r[" & Sheet1.Cells(x, 11).Value - 1 & "]c7)/" & Sheet1.Cells(x, 11).Value & "),4)"
Else
Sheet1.Cells(x, 12).FormulaR1C1 = "=r[-1]c12"
End If
End If
Next x
'to update final table
Application.ScreenUpdating = False
Application.Calculation = xlCalculationManual
Dim LastFinalList As Integer: LastFinalList = Sheet1.Cells(Rows.Count, 15).End(xlUp).Row
For Col = 16 To 19
For c = 2 To LastFinalList
searchrange = Sheet1.Range("J:L")
lookfor = Sheet1.Cells(c, 15) & Sheet1.Cells(1, Col)
CountFor = Application.VLookup(lookfor, searchrange, 3, False)
If IsError(CountFor) Then
Sheet1.Cells(c, Col).Value = "0"
Else
Sheet1.Cells(c, Col).Value = CountFor
End If
Next c
Next Col
Application.Calculation = xlCalculationAutomatic
Application.ScreenUpdating = True
Sheet1.Range("O1").Select
End Sub
Also, initially i had a SUMIF formula instead of application.vlookup to be input in each cell in the converted table. But the code was working as slow as now an was bit bulky, that's why i've decide to switch to VLOOKUP.
The thing is, if it actually the way application.vlookup works (with 0.3sec delay for each row), then i guess there's nothing that can be done, and i'm ok to accept that. Although, if that's not the case, i'd really appreciate if someone could help me out and speed up the process.
Thanks!
You can redefine your LastFinalList variable something like
LastFinalList = Sheets("Sheet1").UsedRange.Rows.Count
OR
LastFinalList = Sheets("Sheet1").Cells(Rows.Count, 2).End(xlUp).Row
instead of explicitly defining used range.
Also use following line of code before your code
Application.ScreenUpdating = False
(Turn screen updating off to speed up macro code. User won't be able to see what the macro is doing, but it will run faster.)
After the whole code runs you can(optional) turn screen updating on with
Application.ScreenUpdating = True
It appears that application.vlookup in my particular case was indeed working very slow (no idea why, though). I've managed to improve macro by replacing vlookup with SUMIF formula in each cell, so now the converted table is updated instantly. Thanks everyone who participated and provided their suggestions!
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I have a method from a long script that creates a hash from genetic sequences, however it is really messy and thus I was wondering whether there was a way to put it more elegantly.
Here is a sample of the script (i.e. it contains an example)...
def make_hash(motif)
main_hash = Hash.new
id = ">isotig00009_f2_3 ~: S.P. Cleavage Site: 22:23 - S.P. D-value: 0.532"
seq = "MLKCFSIIMGLILLLEIGGGCA~IYFYRAQIQAQFQKSLTDVTITDYRENADFQDLIDALQSGLSCCGVNSYEDWDNNIYFNCSGPANNPEALWCAFLLLYTGSSKRSSQHPVRLWSSFPRTTKYFPHKDLHHWLCGYVYNVD"
id_hash = Hash[[[:id_start, :id_end], id.split("~").map(&:strip)].transpose]
seq_hash = Hash[[[:signalp, :seq_end], seq.split("~").map(&:strip)].transpose]
signalp = seq_hash[:signalp]
new_seq_end = seq_hash[:seq_end].gsub(/#{motif}/, '<span class="motif">\0</span>')
new_seq_hash = Hash[:signalp => signalp, :new_seq_end => new_seq_end ]
main_hash[id_hash] = [new_seq_hash]
return main_hash
end
motif = "VT|QAQ|F.D"
main_hash = make_hash(motif)
main_hash.each do |id_hash, seq_hash|
puts id_hash[:id_start]
puts id_hash[:id_end]
puts seq_hash[0][:signalp]
puts seq_hash[0][:new_seq_end]
end
So Is there a more elegant way to write the make_hash method...
Many Thanks
I haven't tested this, but I think this simplification will work:
def make_hash(motif)
id = ">isotig00009_f2_3 ~: S.P. Cleavage Site: 22:23 - S.P. D-value: 0.532"
seq = "MLKCFSIIMGLILLLEIGGGCA~IYFYRAQIQAQFQKSLTDVTITDYRENADFQDLIDALQSGLSCCGVNSYEDWDNNIYFNCSGPANNPEALWCAFLLLYTGSSKRSSQHPVRLWSSFPRTTKYFPHKDLHHWLCGYVYNVD"
id_hash = Hash[[[:id_start, :id_end], id.split("~").map(&:strip)].transpose]
f, s = seq.split("~").map(&:strip)
s.gsub!(/#{motif}/, '<span class="motif">\0</span>')
new_seq_hash = Hash[Hash[:signalp, f], Hash[:new_seq_end, s]]
Hash[id_hash, new_seq_hash]
end
If (as it appears) id and seq both have constant values, you might consider breaking them apart manually, rather than with id.split("~").map(&:strip); i.e.,
id1 = ">isotig00009_f2_3
id2 = ": S.P. Cleavage Site: 22:23 - S.P. D-value: 0.532"
seq1 = "MLKCFSIIMGLILLLEIGGGCA"
seq2 = "IYFYRAQIQAQFQKSLTDVTITDYRENADFQDLIDALQSGLSCCGVNSYEDWDNNIYFNCSGPANNPEALWCAFLLLYTGSSKRSSQHPVRLWSSFPRTTKYFPHKDLHHWLCGYVYNVD"
If there were a need to make seq2 more readable, we could use the "line continuation" character, \ (which even works within strings) like this:
seq2 = "IYFYRAQIQAQFQKSLTDVTITDYRENADFQDLIDALQSGLSCCGVNSYEDWDNNIYFNC"\
"SGPANNPEALWCAFLLLYTGSSKRSSQHPVRLWSSFPRTTKYFPHKDLHHWLCGYVYNVD"
or this:
seq2 = "IYFYRAQIQAQFQKSLTDVTITDYRENADFQDLIDALQSGLSCCGVNSYEDWDNNIYFNC\
SGPANNPEALWCAFLLLYTGSSKRSSQHPVRLWSSFPRTTKYFPHKDLHHWLCGYVYNVD"
If you preferred, you could make 'id' and 'seq' constants ('ID' and 'SEQ', say) and move them outside the method definition. Not surprisingly, line continuation also works for constant strings.
I have a ruby multi-line string (called efixes) that looks like:
ID STATE LABEL INSTALL TIME UPDATED BY ABSTRACT
=== ===== ========== ================= ========== ======================================
1 S hayo32.02 xxxxxxx xxxxxxxx xxxxxxxxxxxxxxx
2 S 23434.23 xxxxxxx xxxxxxxx xxxxxxxxxxxxxxx
STATE codes:
S = STABLE
M = MOUNTED
U = UNMOUNTED
Q = REBOOT REQUIRED
B = BROKEN
I = INSTALLING
R = REMOVING
T = TESTED
P = PATCHED
N = NOT PATCHED
SP = STABLE + PATCHED
SN = STABLE + NOT PATCHED
QP = BOOT IMAGE MODIFIED + PATCHED
QN = BOOT IMAGE MODIFIED + NOT PATCHED
RQ = REMOVING + REBOOT REQUIRED
I only want to display the lines that start with a number. I am having trouble, it doesn't seem to be matching. I found this solution here, (that I don't truly understand right now):
efixes_array = efixes.split("\n").select{|x| /\A[0-9]/.match(x)}
io.puts efixes_array.collect{|x| x.scan(/\A[0-9]/)}.flatten
It is only matching the numbers. I want to display the entire line. The end result, I want to display what is under the "LABELS" column.
This line from your example code
efixes.split("\n").select{|x| /\A[0-9]/.match(x)}
returns an array with all lines that start with a number.