** Python Support Thread ** - Page 5 - Computer Technical Help

going back to the "don't mess with an iterable while you are iterating through it",

it is perfectly OK (read: safe and not alarming) to do this, right?

Code:

for key, value in somedict.iteritems():
    somedict[key] = somefunction(value)

or (if we have mutable types as values):

Code:

for key, value in somedict.iteritems():
    value.somealteringfunction(key, value)

Quote

06-19-2011 , 10:45 PM

#102

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

At first, I was doing something similar to:

Code:

while....:
    a = x%y
    if a == 0:
        .....
    else:
        print x

which I guess is the same thing of sorts and serves as a better visual to why the 'test' primitive was in the block instead of outside it.

Not sure how you program without the else blocks. Are you saying you don't specify else or you really don't use it at all? I'm sort of in the habit of using indenting to execute the else opposed to actually writing else. But there are times it is handy. The next two projects are 'brute force algorithms,' which require some else blocks to execute I think, if for nothing else but to return Null values to the main functions. But, eh, I haven't put any effort into much more than a rudimentary start. Hopefully, this one won't take a week to do.

Quote

06-19-2011 , 11:01 PM

#103

Neil S

King of the sidebar

Join Date: Sep 2004 Posts: 19,637

Spoilers won't hide the answer from bots by the way. It doesn't do a live call toget the spoiler contents. It just hides or shows a div that's already there.

Quote

06-19-2011 , 11:09 PM

#104

sorrow

old hand

Join Date: Apr 2008 Posts: 1,531

Quote:

Originally Posted by RoundTower

going back to the "don't mess with an iterable while you are iterating through it",

it is perfectly OK (read: safe and not alarming) to do this, right?

Code:

for key, value in somedict.iteritems():
    somedict[key] = somefunction(value)

or (if we have mutable types as values):

Code:

for key, value in somedict.iteritems():
    value.somealteringfunction(key, value)

Good question, i've always assumed yes. I generally iterate on the key rather than value though.

Quote

06-19-2011 , 11:14 PM

#105

Neil S

King of the sidebar

Join Date: Sep 2004 Posts: 19,637

Yeah that should be fine.

But you don't want to be adding or deleting. That's typically going to be undefined or weird.

Quote

06-20-2011 , 07:53 PM

#106

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

This code does exactly what I want it to do, but I wonder if there is a simpler solution to this problem. I'm also wondering if there is a different solution to this problem that I didn't catch.

The idea is to find all the values of n via n = 6a + 9b + 20c. I created an ascending list with no repeating values. The program runs pretty slow, but I guess that isn't too surprising since I am running 3 simultaneous loops, checking vs an if, creating a list, sorting the list, then minimizing the list.

Considering the problem set, I could use a smaller value than 150. I ran it with 50 as well, but it is still slow.

Code:

nList = []
for a in range(0,150):
    for b in range(0,150):
        for c in range(0,150):
            n = 6*a + 9*b + 20*c
            if n < 150:
                nList.append(n)
                nList.sort()
                nList = list(set(nList))
print nList

edit to add:

ldo, math ftl?

Code:

nList = []
for a in range(0,10):
    for b in range(0,6):
        for c in range(0,3):
            n = 6*a + 9*b + 20*c
            if n < 50:
                nList.append(n)
                nList.sort()
                nList = list(set(nList))
print nList

Last edited by daveT; 06-20-2011 at 08:05 PM. Reason: now, it's lightening fast.

Quote

06-20-2011 , 08:11 PM

#107

tyler_cracker

Carpal \'Tunnel

Join Date: Apr 2005 Posts: 15,828

why are you sorting and uniq'ing during every iteration?

Quote

06-20-2011 , 08:40 PM

#108

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

Quote:

Originally Posted by tyler_cracker

why are you sorting and uniq'ing during every iteration?

I have to check if n, n + 1,... n+5 is possible. According to the theorem, once I have that, then all n + (inf) is possible. I know that the highest possible number n that can't work is 37.

[0, 6, 9, 12, 15, 18, 20, 21, 24, 26, 27, 29, 30, 32, 33, 35, 36, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69]

The plan (for now) is to check all numbers in the list against n, n + 1,... n+5 (which probably creates another for loop somewhere else), then print out 37.

Quote

06-20-2011 , 08:57 PM

#109

RoundTower

ɹǝʍoʇpunoɹ

Join Date: Feb 2005 Posts: 14,563

that's all gobbledegook to me but even if you really, really need to call

Code:

nList = list(set(nList))

on every iteration, which I'm sure you don't, the expensive sort() operation in the previous line changes absolutely nothing because set() unsorts the list. But really what you should do is make nList a set from the start and call nlist.add(n) instead of your 3 lines.

Quote

06-20-2011 , 09:31 PM

#110

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

here is the problem set that explains what I am doing.

http://ocw.mit.edu/courses/electrica...ents/pset2.pdf

Quote

06-20-2011 , 09:32 PM

#111

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

Quote:

Originally Posted by RoundTower

that's all gobbledegook to me but even if you really, really need to call

Code:

nList = list(set(nList))

How hard is it for me to highlight and format >> comment to see what happens without sort()?

But, yes, this is good. Thanks.

Quote

06-20-2011 , 11:23 PM

#112

Giant Tortoise

centurion

Join Date: Apr 2008 Posts: 171

Quote:

Originally Posted by daveT

I'm currently working through the OCW Intro to Programming course too. I just did the McNugget problem last week. I'm obviously a Python n00b as well, so take my advice with a grain of salt, but what you're trying to do with lists seems needlessly complicated.

I solved this one pretty easily just brute-forcing the possible combinations and using if/else statements to set the appropriate flags and keep track of a counting variable until I found 6 valid combos in a row.

The code is in a spoiler below in case you don't want to see before you solve it for yourself :

Spoiler:

Quote

06-21-2011 , 02:19 AM

#113

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

I have a solution, though it probably isn't as correct as yours. This is something I threw together over about 4 hours (which is extremely fast for me). I got a little to married to the list idea, but I'll work on a different solution at a later date.

Spoiler:

Quote

06-21-2011 , 08:23 AM

#114

jjshabado

Carpal Tunnel

Join Date: Jul 2006 Posts: 22,732

What's this for:

Quote:

list(set(nList))

You should very, very rarely need to construct two data structures like this. You definitely shouldn't do it multiple times. In your case, why wouldn't you just make nList and noValue set's to start?

In fact: I think you can just leave them as sets and then call max(yourSet) to get the largest number.

Edit: The reason I'm pointing that out is that learning to use the right data structure and avoiding lots of unnecessary conversions between data structures is pretty important. Much more important than trying to get code down to a small size or to figure out the very best algorithm to use.

Quote

06-21-2011 , 08:56 AM

#115

Slick Strawberry

journeyman

Join Date: Nov 2010 Posts: 349

Been trying to learn python for a few months on and off in the free time that I have (very little). This week it finally paid off as I could write a script that eases some of the work I was doing.
Even though it is probably a horrible piece of code from a programmers' view, I was still kinda proud. It's 8 times a similar thing after each other, but each one is dependent on the step before it, so I split it into 8 codeblocks. Is there a way to do that easier?
Rough example: I have x, out of that comes x1 x2 and x3, then I do roughly the same with those and get y11 y12 y13, y21 y22 and y23, and y31/y32/y33.
Is that possible to do in one codeblock without using so many for and if blocks that I tab out of the right side of my screen?

Attached the script btw, some variable names might be weird but they make sense if you know what the subject is about. It basically gets the relevant information out of (in this case) 8 documents and pastes it exactly in the format I want in a new file.

Also, first time dabbling in regexes so it took me a while to figure out all those to get exactly what I wanted.

Spoiler:

Code:

#! /usr/local/bin/python

import re
import os

inputfile='' 
path='C:\\blah\\blah\\'
filelisting=os.listdir(path)
for file in filelisting:
    if file[-3:] == 'dot':
        input=open(path+file)
        inputfile=inputfile+input.read()
        
result=open("C:\\blah\\blah\\results.txt", "w")
result.write('digraph G  {\n')

templinklistresult=open("C:\\blah\\blah\\tempresults.txt", "w")

rank1parents=[]     #store all parents for later use
rank2parents=[]
rank3parents=[]
rank4parents=[]
rank5parents=[]
rank6parents=[]
rank7parents=[]
rank8parents=[]

ideal='000101101110101101'                          #first find ideotype
regex = r'([01]+)\s->\s('+str(ideal)+')(.+)'             #make regex with ideotype
ideotypelinks=re.findall(regex,inputfile)               #do the search


ideoparents=[]
for x in xrange(0,len(ideotypelinks)):    #go through all found ideoparents
    if ideotypelinks[x][0] not in ideoparents: #if-block to remove duplicates
        ideoparents.append(ideotypelinks[x][0])
        templinklistresult.write(ideotypelinks[x][0]+' -> '+ideotypelinks[x][1]+ideotypelinks[x][2]+'\n') #write links to file
        rank1parents.append(ideotypelinks[x][0])
        
        
templinklistresult.write('\n2nd rank\n')        #next generation (one up)

ideoparents1=[]
parents=[]
for x in xrange(0,len(ideoparents)):
    searchword=r'([01]+)\s->\s('+ideoparents[x]+')(.+)' #make regex for each parent
    linklist=re.findall(searchword,inputfile)   #make the list of links/parent
    tempparentslist=[]
    for x in xrange(0,len(linklist)):
        if linklist[x][0] not in parents: #if-block to store parents for
            parents.append(linklist[x][0]) #next generation search start
        if linklist[x][0] not in tempparentslist:   #if-block to remove dups
            tempparentslist.append(linklist[x][0])
            templinklistresult.write(linklist[x][0])    #write all three parts of link 
            templinklistresult.write(' -> ')            #to the file with a newline each 
            templinklistresult.write(linklist[x][1])   #time
            templinklistresult.write(linklist[x][2]+'\n')   
            rank2parents.append(linklist[x][0])
    templinklistresult.write('\n')
    
templinklistresult.write('3rd rank\n')      #next generation (one up)
parentslist=[]
for x in xrange(0,len(parents)):
    searchword=r'([01]+)\s->\s('+parents[x]+')(.+)' #make regex for each parent
    linklist=re.findall(searchword,inputfile)   #make the list of links/parent
    tempparentslist=[]
    for x in xrange(0,len(linklist)):
        if linklist[x][0] not in parentslist: #if-block to store parents for
            parentslist.append(linklist[x][0]) #next generation search start
        if linklist[x][0] not in tempparentslist:   #if-block to remove dups
            tempparentslist.append(linklist[x][0])
            templinklistresult.write(linklist[x][0])    #write all three parts of link 
            templinklistresult.write(' -> ')            #to the file with a newline each 
            templinklistresult.write(linklist[x][1])   #time
            templinklistresult.write(linklist[x][2]+'\n')   
            rank3parents.append(linklist[x][0])
    templinklistresult.write('\n')

templinklistresult.write('4th rank\n')      #next generation (one up)
parents=[]
for x in xrange(0,len(parentslist)):
    searchword=r'([01]+)\s->\s('+parentslist[x]+')(.+)' #make regex for each parent
    linklist=re.findall(searchword,inputfile)   #make the list of links/parent
    tempparentslist=[]
    for x in xrange(0,len(linklist)):
        if linklist[x][0] not in parents: #if-block to store parents for
            parents.append(linklist[x][0]) #next generation search start
        if linklist[x][0] not in tempparentslist:   #if-block to remove dups
            tempparentslist.append(linklist[x][0])
            templinklistresult.write(linklist[x][0])    #write all three parts of link 
            templinklistresult.write(' -> ')            #to the file with a newline each 
            templinklistresult.write(linklist[x][1])   #time
            templinklistresult.write(linklist[x][2]+'\n')   
            rank4parents.append(linklist[x][0])
    templinklistresult.write('\n')

templinklistresult.write('5th rank\n')      #next generation (one up)
parentslist=[]
for x in xrange(0,len(parents)):
    searchword=r'([01]+)\s->\s('+parents[x]+')(.+)' #make regex for each parent
    linklist=re.findall(searchword,inputfile)   #make the list of links/parent
    tempparentslist=[]
    for x in xrange(0,len(linklist)):
        if linklist[x][0] not in parentslist: #if-block to store parents for
            parentslist.append(linklist[x][0]) #next generation search start
        if linklist[x][0] not in tempparentslist:   #if-block to remove dups
            tempparentslist.append(linklist[x][0])
            templinklistresult.write(linklist[x][0])    #write all three parts of link 
            templinklistresult.write(' -> ')            #to the file with a newline each 
            templinklistresult.write(linklist[x][1])   #time
            templinklistresult.write(linklist[x][2]+'\n')   
            rank5parents.append(linklist[x][0])
    templinklistresult.write('\n')
    
templinklistresult.write('6th rank\n')      #next generation (one up)
parents=[]
for x in xrange(0,len(parentslist)):
    searchword=r'([01]+)\s->\s('+parentslist[x]+')(.+)' #make regex for each parent
    linklist=re.findall(searchword,inputfile)   #make the list of links/parent
    tempparentslist=[]
    for x in xrange(0,len(linklist)):
        if linklist[x][0] not in parents: #if-block to store parents for
            parents.append(linklist[x][0]) #next generation search start
        if linklist[x][0] not in tempparentslist:   #if-block to remove dups
            tempparentslist.append(linklist[x][0])
            templinklistresult.write(linklist[x][0])    #write all three parts of link 
            templinklistresult.write(' -> ')            #to the file with a newline each 
            templinklistresult.write(linklist[x][1])   #time
            templinklistresult.write(linklist[x][2]+'\n')   
            rank6parents.append(linklist[x][0])
    templinklistresult.write('\n')   


templinklistresult.write('7th rank\n')      #next generation (one up)
parentslist=[]
for x in xrange(0,len(parents)):
    searchword=r'([01]+)\s->\s('+parents[x]+')(.+)' #make regex for each parent
    linklist=re.findall(searchword,inputfile)   #make the list of links/parent
    tempparentslist=[]
    for x in xrange(0,len(linklist)):
        if linklist[x][0] not in parentslist: #if-block to store parents for
            parentslist.append(linklist[x][0]) #next generation search start
        if linklist[x][0] not in tempparentslist:   #if-block to remove dups
            tempparentslist.append(linklist[x][0])
            templinklistresult.write(linklist[x][0])    #write all three parts of link 
            templinklistresult.write(' -> ')            #to the file with a newline each 
            templinklistresult.write(linklist[x][1])   #time
            templinklistresult.write(linklist[x][2]+'\n')   
            rank7parents.append(linklist[x][0])
    templinklistresult.write('\n')    

templinklistresult.write('8th rank\n')      #next generation (one up)
parents=[]
for x in xrange(0,len(parentslist)):
    searchword=r'([01]+)\s->\s('+parentslist[x]+')(.+)' #make regex for each parent
    linklist=re.findall(searchword,inputfile)   #make the list of links/parent
    tempparentslist=[]
    for x in xrange(0,len(linklist)):
        if linklist[x][0] not in parents: #if-block to store parents for
            parents.append(linklist[x][0]) #next generation search start
        if linklist[x][0] not in tempparentslist:   #if-block to remove dups
            tempparentslist.append(linklist[x][0])
            templinklistresult.write(linklist[x][0])    #write all three parts of link 
            templinklistresult.write(' -> ')            #to the file with a newline each 
            templinklistresult.write(linklist[x][1])   #time
            templinklistresult.write(linklist[x][2]+'\n')   
            rank8parents.append(linklist[x][0])
    templinklistresult.write('\n') 


allparentslist=[]   #to not have any duplicates in parentlist

if len(rank8parents) != 0:
    result.write('''{
		rank = same;\n\t\t''')
for x in xrange(0,len(rank8parents)):
    if rank8parents[x] not in allparentslist:
        allparentslist.append(rank8parents[x])
        searchword=rank8parents[x]+'[^;]*;' #everything up to final /table end
        table=re.search(searchword, inputfile)
        result.write(table.group()+'\n\t\t')

if len(rank8parents) != 0:
    result.write('''}\n\t''')
        
if len(rank7parents) != 0:
    result.write('''{
		rank = same;\n\t\t''')
for x in xrange(0,len(rank7parents)):
    if rank7parents[x] not in allparentslist:
        allparentslist.append(rank7parents[x])
        searchword=rank7parents[x]+'[^;]*;' #everything up to final /table end
        table=re.search(searchword, inputfile)
        result.write(table.group()+'\n\t\t')
        
result.write('''}
	''')
        
if len(rank6parents) != 0:
    result.write('''{
		rank = same;\n\t\t''')
for x in xrange(0,len(rank6parents)):
    if rank6parents[x] not in allparentslist:
        allparentslist.append(rank6parents[x])
        searchword=rank6parents[x]+'[^;]*;' #everything up to final /table end
        table=re.search(searchword, inputfile)
        result.write(table.group()+'\n\t\t')
        
        
result.write('''}
	''')
        
if len(rank5parents) != 0:
    result.write('''{
		rank = same;\n\t\t''')
for x in xrange(0,len(rank5parents)):
    if rank5parents[x] not in allparentslist:
        allparentslist.append(rank5parents[x])
        searchword=rank5parents[x]+'[^;]*;' #everything up to final /table end
        table=re.search(searchword, inputfile)
        result.write(table.group()+'\n\t\t')
        
        
result.write('''}
	''')
        
if len(rank4parents) != 0:
    result.write('''{
		rank = same;\n\t\t''')
for x in xrange(0,len(rank4parents)):
    if rank4parents[x] not in allparentslist:
        allparentslist.append(rank4parents[x])
        searchword=rank4parents[x]+'[^;]*;' #everything up to final /table end
        table=re.search(searchword, inputfile)
        result.write(table.group()+'\n\t\t')
        
        
result.write('''}
	''')
        
if len(rank3parents) != 0:
    result.write('''{
		rank = same;\n\t\t''')
for x in xrange(0,len(rank3parents)):
    if rank3parents[x] not in allparentslist:
        allparentslist.append(rank3parents[x])
        searchword=rank3parents[x]+'[^;]*;' #everything up to final /table end
        table=re.search(searchword, inputfile)
        result.write(table.group()+'\n\t\t')
        
        
result.write('''}
	''')
        
if len(rank2parents) != 0:
    result.write('''{
		rank = same;\n\t\t''')
for x in xrange(0,len(rank2parents)):
    if rank2parents[x] not in allparentslist:
        allparentslist.append(rank2parents[x])
        searchword=rank2parents[x]+'[^;]*;' #everything up to final /table end
        table=re.search(searchword, inputfile)
        result.write(table.group()+'\n\t\t')
        
        
result.write('''}
	''')
        
if len(rank1parents) != 0:
    result.write('''{
		rank = same;\n\t\t''')
for x in xrange(0,len(rank1parents)):
    if rank1parents[x] not in allparentslist:
        allparentslist.append(rank1parents[x])
        searchword=rank1parents[x]+'[^;]*;' #everything up to final /table end
        table=re.search(searchword, inputfile)
        result.write(table.group()+'\n\t\t')

result.write('''}
	''')        
result.write('''{
		rank = same;\n\t\t''')
    
#paste ideotype table in there
searchword=ideal+'[^;]*;'
table=re.search(searchword, inputfile)
result.write(table.group()+'\n\t\t')


result.write('''Schedulename\n\t\t[shape=egg,fillcolor=gray20,fontcolor=white,fontsize=18,style=filled,label="all"];\n\t}''')

result.write('\n')
templinklistresult.close()
templinklistresult=open("C:\\blah\\blah\\tempresults.txt", "r")
for line in templinklistresult:
    if len(line) >= 10:
        result.write(line)
result.write("}")

Quote

06-22-2011 , 07:40 AM

#116

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

Quote:

Originally Posted by jjshabado

What's this for:

You should very, very rarely need to construct two data structures like this. You definitely shouldn't do it multiple times. In your case, why wouldn't you just make nList and noValue set's to start?

In fact: I think you can just leave them as sets and then call max(yourSet) to get the largest number.

Edit: The reason I'm pointing that out is that learning to use the right data structure and avoiding lots of unnecessary conversions between data structures is pretty important. Much more important than trying to get code down to a small size or to figure out the very best algorithm to use.

Okay, I'll look all this up.

Slick Strawberry: The reason we were using spoilers is so we don't show off the answers to the homework we are doing.

Quote

06-22-2011 , 09:16 AM

#117

Slick Strawberry

journeyman

Join Date: Nov 2010 Posts: 349

I'm using a spoiler because a spoiler gets a scrollbar if the script is too long, don't want my posts to be extremely long and I don't really know a good other way to show off my code.

Quote

06-22-2011 , 12:01 PM

#118

tyler_cracker

Carpal \'Tunnel

Join Date: Apr 2005 Posts: 15,828

Use the [ code ] tag.

Last edited by tyler_cracker; 06-22-2011 at 12:01 PM. Reason: Or continue to use spoilers. Just know there's an option.

Quote

06-22-2011 , 12:15 PM

#119

Slick Strawberry

journeyman

Join Date: Nov 2010 Posts: 349

Didn't know the code tag also added scrollbars to the code, now I have a code tag in a spoiler tag...

Quote

06-22-2011 , 12:39 PM

#120

TheIrishThug

Pooh-Bah

Join Date: Jan 2005 Posts: 5,223

You can also use the php tag to get some syntax highlighting. It isn't perfect because it as it is geared towards php, but it does a decent job with any c-style language.

Quote

06-24-2011 , 01:28 PM

#121

jjshabado

Carpal Tunnel

Join Date: Jul 2006 Posts: 22,732

What's the best way for iterating over a list while filtering out values?

For example assume I have the following:

Code:

l = [1,None,3]
for x in l:
    if x:
         print 'Do a bunch of stuff'

So I can do:

Code:

l = [1,None,3]
for x in [y for y in l if y]:
    print 'Do a bunch of stuff'

But that still seems a bit wonky. I don't think there are any performance implications but it just doesn't look that great.

Any better ideas?

Quote

06-24-2011 , 02:05 PM

#122

RoundTower

ɹǝʍoʇpunoɹ

Join Date: Feb 2005 Posts: 14,563

first one looks fine to me.

You can also use

Code:

for x in filter(None, l):

Quote

06-24-2011 , 02:09 PM

#123

RoundTower

ɹǝʍoʇpunoɹ

Join Date: Feb 2005 Posts: 14,563

in Python 3 I understand you could use

Code:

[print(x) for x i l if x]

which returns a list of None, but has the correct side-effect of printing the non-False values. This won't work in Python 2 because print is not an object. You could wrap whatever you need to do inside the loop in a callable object though.

Quote

06-24-2011 , 02:11 PM

#124

jjshabado

Carpal Tunnel

Join Date: Jul 2006 Posts: 22,732

I was wondering how you'd do it without the use of filter (since some people are against that).

Your second one won't work for me since my print statement is just standing in for about half a dozen statements and I hate putting a lot of logic inside of a list comprehension.

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06-24-2011 , 07:01 PM

#125

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

May seem totally out of place, but the only thing I am against is using 'l' and '1' in the same program. People like to use various fonts and some of those fonts aren't going to show the difference between 'l' and '1'. Even with a decent font, it is still easy to mistaken the stuff.

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