# Embedding if statements in upto function.(newbie question)

**URL:** <https://www.ruby-forum.com/t/embedding-if-statements-in-upto-function-newbie-question/85285>\
**Category:** Ruby\
**Created:** [February 12, 2007, 12:04pm UTC](https://www.ruby-forum.com/t/embedding-if-statements-in-upto-function-newbie-question/85285 "2007-02-12T12:04:58Z")\
**Posts on this page:** 7\
**Page:** 1

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**Author:** ![wickie](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/ba9def/32.png) [@wickie](https://www.ruby-forum.com/u/wickie)\
**Post date:** [February 12, 2007, 12:04pm UTC](https://www.ruby-forum.com/t/embedding-if-statements-in-upto-function-newbie-question/85285/1 "2007-02-12T12:04:58Z")

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Thank you to everyone for all of your help.  
I was trying to use upto to go through an array of arrays while using if  
statments. Could anyone enlighten me on what I am doing wrong with the  
if statements?  
I am sure it is something easy.

I tried this:  
0.upto( a.size-2 ) {|i|  
if (a[i][2]==a[i+1][2] and a[i][0]==a[i+1][0]) then  
print { (a[i+1][3] - a[i][3]) / a[i][3] \* 100 } /n  
else print 0  
end}

Thanks again for all of the help.  
Michael

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**Author:** ![wickie](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/ba9def/32.png) [@wickie](https://www.ruby-forum.com/u/wickie)\
**Post date:** [February 12, 2007, 12:54pm UTC](https://www.ruby-forum.com/t/embedding-if-statements-in-upto-function-newbie-question/85285/2 "2007-02-12T12:54:51Z")

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On Mon, Feb 12, 2007 at 08:05:01PM +0900, Michael Sc wrote:

> else print 0  
> end}

Why don’t you give some sample data, and show what happens when you run  
it.  
Preferably make a small, standalone program which demonstrates the  
problem.  
Do you get an exception raised? A wrong result?

Looking at the above, the fact that you have used curly brackets in the  
‘print’ statement certainly won’t help… this creates a code block and  
will  
give you a confusing error like this:

irb(main):003:0\> print { 4 + 5 } / 3  
nilNoMethodError: undefined method `/’ for nil:NilClass  
from (irb):3  
from :0

This is equivalent to:

print( { 4+5 } ) / 3  
^ ^ ^  
| | |  
| | value returned by  
| | print() is nil  
| |  
| a code block  
| which is not used  
|  
no arguments  
to print

BTW you probably want to use puts rather than print, so your numbers are  
on  
separate lines; or at least print a space between them.

Your above code does also highlight a skeleton in Ruby’s closet: note  
that

```
puts (4+5)/3

puts(4+5)/3

```

are two different constructs, even though they differ only by one space.  
They are equivalent to, respectively,

```
puts( (4+5)/3 ) # probably what you want

( puts(4+5) ) / 3 # prints 9, tries to divide nil by 3

```

B.

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**Author:** ![wickie](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/ba9def/32.png) [@wickie](https://www.ruby-forum.com/u/wickie)\
**Post date:** [February 12, 2007, 1:49pm UTC](https://www.ruby-forum.com/t/embedding-if-statements-in-upto-function-newbie-question/85285/3 "2007-02-12T13:49:51Z")

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Thanks Brian, I appreciate the help.

I changed the code and ran it again.  
0.upto( a.size-2 ) {|i|  
if (a[i][2]==a[i+1][2] and a[i][0]==a[i+1][0]) then  
puts ( (a[i+1][3] - a[i][3]) / a[i][3] \* 100 )  
else puts 0  
end}  
With this I get an error on the 3rd line of undefined method(-) for “.08  
#(the first number)#”:String(NoMethodError)Print was able to do the  
calculation without the if statements. I think I misunderstood how  
output in FasterCSV processes files. basically, I sorted a csv file and  
I was trying to do some simple math on that file.

Basically here is what I need to do. I have very large csv files and I  
wanted to create percentage changes. Should I try to change a to an  
array?

I have a file that looks like this  
x1 date1 y1 #  
x1 date2 y1 #  
…  
x1 date1 y2 #  
x1 date2 y2 #  
…  
x2 date1 y1 #  
x2 date2 y1 #

I just want to returns only if columns 1 and 3 equal the next point in  
the column.  
Thanks again.  
Michael  
Brian C. wrote:

> On Mon, Feb 12, 2007 at 08:05:01PM +0900, Michael Sc wrote:
> 
> > else print 0  
> > end}
> 
> Why don’t you give some sample data, and show what happens when you run  
> it.  
> Preferably make a small, standalone program which demonstrates the  
> problem.  
> Do you get an exception raised? A wrong result?

---

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**Author:** ![wickie](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/ba9def/32.png) [@wickie](https://www.ruby-forum.com/u/wickie)\
**Post date:** [February 12, 2007, 2:56pm UTC](https://www.ruby-forum.com/t/embedding-if-statements-in-upto-function-newbie-question/85285/4 "2007-02-12T14:56:07Z")

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> With this I get an error on the 3rd line of undefined method(-) for “.08  
> #(the first number)#”:String(NoMethodError)Print was able to do the  
> calculation without the if statements.

I expect the problem is that a[x][y] contains a string, not an integer.

You can confirm this by adding

```
puts a.inspect

```

into your program just after you’ve read in the array, and look for  
[“0”, “1”, “2”] instead of [0, 1, 2]

To fix the problem, try adding some .to\_i calls on the elements before  
performing the arithmetic on them.

irb(main):001:0\> a = “9”  
=\> “9”  
irb(main):002:0\> b = “6”  
=\> “6”  
irb(main):003:0\> a - b  
NoMethodError: undefined method -’ for “9”:String  
from (irb):3  
irb(main):004:0\> a.to\_i - b.to\_i  
=\> 3

Notice that the problem you have come across has nothing to with “if”  
statements, nor with them being embedded in the “upto” construct. The  
moral:  
try to avoid premature diagnosis 🙂

Regards,

Brian.

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**Author:** ![wickie](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/ba9def/32.png) [@wickie](https://www.ruby-forum.com/u/wickie)\
**Post date:** [February 12, 2007, 3:26pm UTC](https://www.ruby-forum.com/t/embedding-if-statements-in-upto-function-newbie-question/85285/5 "2007-02-12T15:26:09Z")

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Brian,  
Thank you very much. You are absolutely correct about the problem and  
premature diagnosis.  
Michael  
Brian C. wrote:

> Notice that the problem you have come across has nothing to with “if”  
> statements, nor with them being embedded in the “upto” construct. The  
> moral:  
> try to avoid premature diagnosis 🙂
> 
> Regards,
> 
> Brian.

---

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**Author:** ![wickie](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/ba9def/32.png) [@wickie](https://www.ruby-forum.com/u/wickie)\
**Post date:** [February 12, 2007, 4:35pm UTC](https://www.ruby-forum.com/t/embedding-if-statements-in-upto-function-newbie-question/85285/6 "2007-02-12T16:35:37Z")

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thank you very much. I had an issue with the string. I adjusted the  
code to the following. I knew it had to be something incredibly trivial.  
Thank you all again.

y=0.upto(a.size - 2 ) do |i|  
if (a[i][2]==a[i+1][2] and a[i][0]==a[i+1][0]) then  
puts ((a[i+1][3].to\_f - a[i][3].to\_f) / a[i][3].to\_f \* 100)  
else  
puts 0  
end

Jeremy McAnally wrote:

> Let’s clean your syntax up a little bit first…
> 
> 0.upto(a.size - 2) do |i|  
> if (a[i][2]==a[i+1][2] and a[i][0]==a[i+1][0])  
> print ((a[i+1][3] - a[i][3]) / a[i][3] \* 100) / n  
> else  
> print 0  
> end  
> end
> 
> Looking at this, I (as someone else already stated) would make sure  
> that everything is a number rather than a string. It seems that if  
> you’re having problems with the if statement, that’s probably the  
> case. I would inspect the objects at every iterations (e.g., rather  
> than just printing 0, I would print a[i][2].inspect and  
> a[i+1][2].inspect or whatever). That should give you some insight as  
> to what’s going on.
> 
> –Jeremy
> 
> On 2/12/07, Michael Sc [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote:
> 
> > else print 0  
> > end}
> > 
> > Thanks again for all of the help.  
> > Michael
> > 
> > –  
> > Posted via [http://www.ruby-forum.com/](http://www.ruby-forum.com/).
> 
> –  
> [http://www.jeremymcanally.com/](http://www.jeremymcanally.com/)
> 
> My free Ruby e-book:  
> [http://www.humblelittlerubybook.com/book/](http://www.humblelittlerubybook.com/book/)
> 
> My blogs:  
> [http://www.mrneighborly.com/](http://www.mrneighborly.com/)  
> [http://www.rubyinpractice.com/](http://www.rubyinpractice.com/)

---

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**Author:** ![wickie](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/ba9def/32.png) [@wickie](https://www.ruby-forum.com/u/wickie)\
**Post date:** [February 12, 2007, 4:07pm UTC](https://www.ruby-forum.com/t/embedding-if-statements-in-upto-function-newbie-question/85285/7 "2007-02-12T16:07:20Z")

</div>

Let’s clean your syntax up a little bit first…

0.upto(a.size - 2) do |i|  
if (a[i][2]==a[i+1][2] and a[i][0]==a[i+1][0])  
print ((a[i+1][3] - a[i][3]) / a[i][3] \* 100) / n  
else  
print 0  
end  
end

Looking at this, I (as someone else already stated) would make sure  
that everything is a number rather than a string. It seems that if  
you’re having problems with the if statement, that’s probably the  
case. I would inspect the objects at every iterations (e.g., rather  
than just printing 0, I would print a[i][2].inspect and  
a[i+1][2].inspect or whatever). That should give you some insight as  
to what’s going on.

–Jeremy

On 2/12/07, Michael Sc [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote:

> else print 0  
> end}
> 
> Thanks again for all of the help.  
> Michael
> 
> –  
> Posted via [http://www.ruby-forum.com/](http://www.ruby-forum.com/).

–  
[http://www.jeremymcanally.com/](http://www.jeremymcanally.com/)

My free Ruby e-book:  
[http://www.humblelittlerubybook.com/book/](http://www.humblelittlerubybook.com/book/)

My blogs:

> **[Site not found · DreamHost](http://www.mrneighborly.com/)**
>
> The owner of this domain has not yet uploaded their website.

[http://www.rubyinpractice.com/](http://www.rubyinpractice.com/)