LAE

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Originally posted by: Iain Sanders

Hi Guys,

Please advise how I can add a decimal point to the following strings:

100 so that it reads: 1.00 and

100y100 so that it reads: 1.00y1.00 and

100y100y100 so that it reads: 1.00y1.00y1.00

Many thanks,

Iain

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Originally posted by: ejones

newdata = data.regexSubstitute("100","1.00")

I may need more context to give a good answer.

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Originally posted by: Iain Sanders

Thanks.

Will this replace every occurrence of 100 with 1.00?

Iain

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Originally posted by: ejones

Yes. And also stonysmith just caught me on this. I over complicated it by using one of the regex functions. The replace() function would be much simpler.

newdata = data.replace("100","1.00")

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Originally posted by: Iain Sanders

The data I have is in the data field called "Fixed1"

Here is a summary of the various records I have to start with (shown below)

The data consists of numbers imported as strings, separated by � delimiters.

Each value should have 2 decimal places, but when the data is imported, the decimal place is removed, so that e.g.

1.00�1.00 becomes 100�100,
1.00�0.60 becomes 100�60,
84.00�2.50 becomes 8400�250 etc...

Basically, I want to be able to place the decimal point back into all the data.

Here is a sample of the data imported:

Fixed1
100�100
100�100
100�100
100�60

100�100
100�100
100�100
100�60
100�100
100�100
100�100
100�60
100�100�100
100�100
100�60
100�100
100�100
100�100
100�100
100�100
100�100

100�100
100
100�100

Note: This was originally posted by an inactive account. Content was preserved by moving under an admin account.

Originally posted by: Iain Sanders

The data I have is in the data field called "Fixed1"

Here is a summary of the various records I have to start with (shown below)

The data consists of numbers imported as strings, separated by � delimiters.

Each value should have 2 decimal places, but when the data is imported, the decimal place is removed, so that e.g.

1.00�1.00 becomes 100�100,
1.00�0.60 becomes 100�60,
84.00�2.50 becomes 8400�250 etc...

Basically, I want to be able to place the decimal point back into all the data.

Here is a sample of the data imported:

Fixed1
100�100
100�100
100�100
100�60

100�100
100�100
100�100
100�60
100�100
100�100
100�100
100�60
100�100�100
100�100
100�60
100�100
100�100
100�100
100�100
100�100
100�100

100�100
100
100�100

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Originally posted by: Iain Sanders

Thank you for your response - one last thing, how do I make this replacement valid for any number, and not just "100"?

Originally posted by: ejones
					

Yes.  And also stonysmith just caught me on this.  I over complicated it by using one of the regex functions.  The replace() function would be much simpler.


newdata = data.replace("100","1.00")
			

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Originally posted by: ejones

There is probably a way to do this with regex. But it would be hard to understand and hard to update. Here is how I would do it.

fixed1_Array = Fixed1.split("�")
fixed1_Array_Length = fixed1_Array.len()
iter = 0
new_Fixed1 = ""
while iter < fixed1_Array_Length {
temp = fixed1_Array[iter].int()
temp_value = format("%d.%02d%"
,temp.div(100)
,temp.mod(100)
)
new_Fixed1 = new_Fixed1.joinStrings(temp_value,"�")
iter = iter + 1
}
emit new_Fixed1 as "Fixed1"

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Originally posted by: rboccuzzi

There are different ways you could do this. [EDIT: I worked on this while Ernest posted the above, so here is ANOTHER way ]

I choose to break apart the list, to one element per row, and then reassemble it using an agg. First, I added a RowIndex, so I could reassemble using the Agg correctly, ensuring that only the ones that were separated got joined.

I attached a graph that does this.

Cheers
Rich
Attachments:
example.brg

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Originally posted by: ejones

I like how your version avoids the conversion from string to integer and back to string. This method probably runs faster.

Originally posted by: rboccuzzi
					

There are different ways you could do this. [EDIT: I worked on this while Ernest posted the above, so here is ANOTHER way  ]


I choose to break apart the list, to one element per row, and then reassemble it using an agg.  First, I added a RowIndex, so I could reassemble using the Agg correctly, ensuring that only the ones that were separated got joined.


I attached a graph that does this.


Cheers

Rich
			

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Originally posted by: Iain Sanders

This was really helpful!

The example with the explanation given below worked a treat!

Very straightforward, efficient and effective.

Thank you so much!

Iain

Originally posted by: rboccuzzi
					

There are different ways you could do this. [EDIT: I worked on this while Ernest posted the above, so here is ANOTHER way  ]


I choose to break apart the list, to one element per row, and then reassemble it using an agg.  First, I added a RowIndex, so I could reassemble using the Agg correctly, ensuring that only the ones that were separated got joined.


I attached a graph that does this.


Cheers

Rich