Programming Forums > Script Programming > Python

commodore · May 14th, 2006, 6:47 AM

I was trying to do mrynit's bob, bill and joe thingie, but it doesn't append the last elif person

bob=['hammer', 'saw']
joe=['saw', 'drill', 'leveler']
bill=['hammer', 'leveler']
everyone=[bob, joe, bill]
countOverall=len(bob)+len(joe)+len(bill)
everyoneString=' '.join(bob+joe+bill)

def toolcount(tool):
    count=0
    finding=everyoneString.find(tool)
    while finding!=-1:
        count+=1
        start=finding+1
        finding=everyoneString.find(tool, start)
    return count

def whoowns(tool):
    owners=[]
    if ' '.join(bob).find(tool)!=-1:
        owners.append('bob')
    elif ' '.join(bill).find(tool)!=-1:
        owners.append('bill')
    elif ' '.join(joe).find(tool)!=-1:
        owners.append('joe')
    print owners

whoowns('leveler')

Arevos · May 14th, 2006, 8:45 AM

That's because elif is just an extension of the original if statement (it's short for "else if"). If the first if statement succeeds, successive elif statements are not executed. Use "if" in place of "elif".

Also, don't use find. Use in! The in operator returns true if an element is in a list.

(Toggle Plain Text)

if tool in bob:
    owners.append('bob')

You might also wish to use a dictionary to store all of the data, rather than several lists. A dictionary matches keys to values. Consider the following dictionary:

(Toggle Plain Text)

people = dict(
    bob = ['hammer', 'saw'],
    joe = ['saw', 'drill', 'leveler'],
    bill = ['hammer', 'leveler']
)

This associates a set of strings with a set of lists. You can also write the same dictionary out in an alternative format:

(Toggle Plain Text)

people = {
    'bob' : ['hammer', 'saw'],
    'joe' : ['saw', 'drill', 'leveler'],
    'bill' : ['hammer', 'leveler']
}

You can use a dictionary to pull out information about a single person:

(Toggle Plain Text)

print "Bob's tools:", people['bob']

Or iterate through all the people:

(Toggle Plain Text)

for person, tools in people.items():
   print person, "has", tools

With a dictionary, finding out who owns what can be easily done with a list comprehension:

(Toggle Plain Text)

def whoowns(tool):
    return [person for person, tools in people.items() if tool in tools]

commodore · May 14th, 2006, 10:06 AM

I thought of using dictionaries, but I didn't know they were very useful.

Dietrich · May 18th, 2006, 11:13 AM

Here is a small math problem from Bitwise Magazine that can be solved with the aid of Python:

Quote:

Ask a friend to choose a secret three-digit number (1 to 999). Then ask to divide it by 7 and tell the remainder. Then ask to divide the original number by 11 and again tell the remainder. Now repeat for a final time with 13. You now have three small numbers, and from these you can quickly identify the original three-digit number.

gryfang · May 18th, 2006, 11:07 PM

Also you could use exercises from one language (like C++, Java, Perl) that is popular and may have a lot of test Questions and try coding them in Python. I find that helpful because if you can't figure it out you can look at "their answer" as a hint (because it isn't python code).
Warning: some elements in one language are over simplified in another, while others are overcomplicated. This changes the challenge of the questions in some cases.

DaWei · May 19th, 2006, 12:23 AM

The problem with using exercises from one language to code into another is that you mostly learn to translate, not to use the specific, individual strengths of a language that set it apart. I could take a C exercise that used char arrays heavily and translate it into Python, but I wouldn't be learning Python, and I wouldn't be improving my C. The thing is to take a problem and solve it in differing ways with different languages.

titaniumdecoy · May 19th, 2006, 2:49 AM

Check out USACO. Once you register you are given increasingly difficult problems to solve. You have to submit each program and have it function correctly to move on to the next set of problems. Accepted languages include C, C++, Java, and Pascal. Even the "easiest" problems are difficult and require reasonable programming experience.

gryfang · May 19th, 2006, 10:18 AM

Quote:

Originally Posted by DaWei

The problem with using exercises from one language to code into another is that you mostly learn to translate, not to use the specific, individual strengths of a language that set it apart. I could take a C exercise that used char arrays heavily and translate it into Python, but I wouldn't be learning Python, and I wouldn't be improving my C. The thing is to take a problem and solve it in differing ways with different languages.

That's really what I meant, I just suggested looking at the other language's code if you get stuck and need help solving the problem. There are several answers to a problem.

Another fun source for interested problems is ACM intercollegiate Programming Competition . They give a List of problem sets that are supposed to be answered "quickly", but you can take your sweet time. Just go to the past problems section and the archives and have fun.

As DaWei said it's the PROBLEM you need to solve. And no the programming isn't just C/C++ it's C/C++ Java, C/C++, and Pascal. But the problems are there so you can solve them in ANY language you want.

I have to make it know that I and for solving problems, PROBLEMS, PROBLEMS, PROBLEMS, not translations (unless that is the problem

).

commodore · May 20th, 2006, 3:21 PM

I didn't code for some time but now I finished the handymans task in a few minutes:

(Toggle Plain Text)

bob=['hammer', 'saw']
joe=['saw', 'drill', 'leveler']
bill=['hammer', 'leveler']
everyone=[bob, joe, bill]
countOverall=len(bob)+len(joe)+len(bill)
everyoneString=' '.join(bob+joe+bill)

def whoowns(tool):
    owners=[]
    if ' '.join(bob).find(tool)!=-1:
        owners.append('bob')
    if ' '.join(bill).find(tool)!=-1:
        owners.append('bill')
    if ' '.join(joe).find(tool)!=-1:
        owners.append('joe')
    return owners

print 'There are ', countOverall, 'tools: \n'
print '%s saws, owned by %s' % (len(whoowns('saw')),', '.join(whoowns('saw')))
print '%s drills, owned by %s' % (len(whoowns('drill')),', '.join(whoowns('drill')))
print '%s levelers, owned by %s' % (len(whoowns('leveler')),', '.join(whoowns('leveler')))
print '%s hammers, owned by %s' % (len(whoowns('hammer')),', '.join(whoowns('hammer')))

Basu · May 22nd, 2006, 10:54 AM

Don't write programs, practice creating your own data types. Create classes that have the properties of things like vectors, complex numbers, Cartesian co-ordinates etc. Take your inspiration from math and physics. While making the classes include functions that properly manipulate and allow calculations using those data types. And if you feel like a real challenge,try writing programs which make use of and convert between multiple such classes. Have fun!

May 14th, 2006, 6:47 AM	#21
commodore Programmer Join Date: Nov 2005 Location: Estonia Posts: 97 Rep Power: 0	I was trying to do mrynit's bob, bill and joe thingie, but it doesn't append the last elif person (Toggle Plain Text) bob=['hammer', 'saw'] joe=['saw', 'drill', 'leveler'] bill=['hammer', 'leveler'] everyone=[bob, joe, bill] countOverall=len(bob)+len(joe)+len(bill) everyoneString=' '.join(bob+joe+bill) def toolcount(tool): count=0 finding=everyoneString.find(tool) while finding!=-1: count+=1 start=finding+1 finding=everyoneString.find(tool, start) return count def whoowns(tool): owners=[] if ' '.join(bob).find(tool)!=-1: owners.append('bob') elif ' '.join(bill).find(tool)!=-1: owners.append('bill') elif ' '.join(joe).find(tool)!=-1: owners.append('joe') print owners whoowns('leveler') bob=['hammer', 'saw'] joe=['saw', 'drill', 'leveler'] bill=['hammer', 'leveler'] everyone=[bob, joe, bill] countOverall=len(bob)+len(joe)+len(bill) everyoneString=' '.join(bob+joe+bill) def toolcount(tool): count=0 finding=everyoneString.find(tool) while finding!=-1: count+=1 start=finding+1 finding=everyoneString.find(tool, start) return count def whoowns(tool): owners=[] if ' '.join(bob).find(tool)!=-1: owners.append('bob') elif ' '.join(bill).find(tool)!=-1: owners.append('bill') elif ' '.join(joe).find(tool)!=-1: owners.append('joe') print owners whoowns('leveler')

May 14th, 2006, 8:45 AM	#22
Arevos Programming Guru Join Date: Aug 2005 Location: England Posts: 1,499 Rep Power: 5	That's because elif is just an extension of the original if statement (it's short for "else if"). If the first if statement succeeds, successive elif statements are not executed. Use "if" in place of "elif". Also, don't use find. Use in! The in operator returns true if an element is in a list. (Toggle Plain Text) if tool in bob: owners.append('bob') if tool in bob: owners.append('bob') You might also wish to use a dictionary to store all of the data, rather than several lists. A dictionary matches keys to values. Consider the following dictionary: (Toggle Plain Text) people = dict( bob = ['hammer', 'saw'], joe = ['saw', 'drill', 'leveler'], bill = ['hammer', 'leveler'] ) people = dict( bob = ['hammer', 'saw'], joe = ['saw', 'drill', 'leveler'], bill = ['hammer', 'leveler'] ) This associates a set of strings with a set of lists. You can also write the same dictionary out in an alternative format: (Toggle Plain Text) people = { 'bob' : ['hammer', 'saw'], 'joe' : ['saw', 'drill', 'leveler'], 'bill' : ['hammer', 'leveler'] } people = { 'bob' : ['hammer', 'saw'], 'joe' : ['saw', 'drill', 'leveler'], 'bill' : ['hammer', 'leveler'] } You can use a dictionary to pull out information about a single person: (Toggle Plain Text) print "Bob's tools:", people['bob'] print "Bob's tools:", people['bob'] Or iterate through all the people: (Toggle Plain Text) for person, tools in people.items(): print person, "has", tools for person, tools in people.items(): print person, "has", tools With a dictionary, finding out who owns what can be easily done with a list comprehension: (Toggle Plain Text) def whoowns(tool): return [person for person, tools in people.items() if tool in tools] def whoowns(tool): return [person for person, tools in people.items() if tool in tools]

May 18th, 2006, 11:07 PM	#25
gryfang Programmer Join Date: May 2006 Location: Ohio Posts: 36 Rep Power: 0	Also you could use exercises from one language (like C++, Java, Perl) that is popular and may have a lot of test Questions and try coding them in Python. I find that helpful because if you can't figure it out you can look at "their answer" as a hint (because it isn't python code). Warning: some elements in one language are over simplified in another, while others are overcomplicated. This changes the challenge of the questions in some cases. __________________ -------------------- LOAD "*" ,8,1 God bless - Gryfang

May 19th, 2006, 12:23 AM	#26
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	The problem with using exercises from one language to code into another is that you mostly learn to translate, not to use the specific, individual strengths of a language that set it apart. I could take a C exercise that used char arrays heavily and translate it into Python, but I wouldn't be learning Python, and I wouldn't be improving my C. The thing is to take a problem and solve it in differing ways with different languages. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

May 20th, 2006, 3:21 PM	#29
commodore Programmer Join Date: Nov 2005 Location: Estonia Posts: 97 Rep Power: 0	I didn't code for some time but now I finished the handymans task in a few minutes: (Toggle Plain Text) bob=['hammer', 'saw'] joe=['saw', 'drill', 'leveler'] bill=['hammer', 'leveler'] everyone=[bob, joe, bill] countOverall=len(bob)+len(joe)+len(bill) everyoneString=' '.join(bob+joe+bill) def whoowns(tool): owners=[] if ' '.join(bob).find(tool)!=-1: owners.append('bob') if ' '.join(bill).find(tool)!=-1: owners.append('bill') if ' '.join(joe).find(tool)!=-1: owners.append('joe') return owners print 'There are ', countOverall, 'tools: \n' print '%s saws, owned by %s' % (len(whoowns('saw')),', '.join(whoowns('saw'))) print '%s drills, owned by %s' % (len(whoowns('drill')),', '.join(whoowns('drill'))) print '%s levelers, owned by %s' % (len(whoowns('leveler')),', '.join(whoowns('leveler'))) print '%s hammers, owned by %s' % (len(whoowns('hammer')),', '.join(whoowns('hammer'))) bob=['hammer', 'saw'] joe=['saw', 'drill', 'leveler'] bill=['hammer', 'leveler'] everyone=[bob, joe, bill] countOverall=len(bob)+len(joe)+len(bill) everyoneString=' '.join(bob+joe+bill) def whoowns(tool): owners=[] if ' '.join(bob).find(tool)!=-1: owners.append('bob') if ' '.join(bill).find(tool)!=-1: owners.append('bill') if ' '.join(joe).find(tool)!=-1: owners.append('joe') return owners print 'There are ', countOverall, 'tools: \n' print '%s saws, owned by %s' % (len(whoowns('saw')),', '.join(whoowns('saw'))) print '%s drills, owned by %s' % (len(whoowns('drill')),', '.join(whoowns('drill'))) print '%s levelers, owned by %s' % (len(whoowns('leveler')),', '.join(whoowns('leveler'))) print '%s hammers, owned by %s' % (len(whoowns('hammer')),', '.join(whoowns('hammer')))

May 14th, 2006, 10:06 AM	#23
commodore Programmer Join Date: Nov 2005 Location: Estonia Posts: 97 Rep Power: 0	I thought of using dictionaries, but I didn't know they were very useful.

May 19th, 2006, 2:49 AM	#27
titaniumdecoy Expert Programmer Join Date: Nov 2005 Posts: 903 Rep Power: 3	Check out USACO. Once you register you are given increasingly difficult problems to solve. You have to submit each program and have it function correctly to move on to the next set of problems. Accepted languages include C, C++, Java, and Pascal. Even the "easiest" problems are difficult and require reasonable programming experience.

May 22nd, 2006, 10:54 AM	#30
Basu Newbie Join Date: May 2006 Posts: 1 Rep Power: 0	Don't write programs, practice creating your own data types. Create classes that have the properties of things like vectors, complex numbers, Cartesian co-ordinates etc. Take your inspiration from math and physics. While making the classes include functions that properly manipulate and allow calculations using those data types. And if you feel like a real challenge,try writing programs which make use of and convert between multiple such classes. Have fun!

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Programming Forums > Script Programming > Python

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