[R] separate numbers from chars in a string

[Marc Schwartz]

On Jul 31, 2014, at 3:17 AM, Uwe Ligges <ligges at statistik.tu-dortmund.de> wrote:

> 
> 
> On 31.07.2014 04:46, carol white wrote:
>> There are some level of variation either chars followed by numbers or chars, numbers, chars
>> 
>> 
>> Perhaps, I should use gsub as you suggested all and if the string is composed of chars followed by numbers, it will return the 3rd part empty?
> 
> 
> Please read about regularvexpressions and describe your problem accurately. If the last strings are onot always present, use * rather than + at the very end of the regular expression.
> 
> Best,
> Uwe Ligges


Carol,

As Uwe notes, reviewing the documentation for ?regex and the examples in ?gsub can be helpful. There are also online regex resources such as:

  http://www.regular-expressions.info

The question is how much variation might be present. If it will always be up to 3 possible components, then as Uwe indicated, using the '*' instead of '+' will allow for the possibility that one or more patterns will not be present. '*' means that 0 or more of the patterns must be present, whereas '+' requires that at least one or more matches are present.

> strsplit(gsub("([a-z]*)([0-9]*)([a-z]*)", "\\1 \\2 \\3", "absdfds0213451ab"), " ")
[[1]]
[1] "absdfds" "0213451" "ab"   

> strsplit(gsub("([a-z]*)([0-9]*)([a-z]*)", "\\1 \\2 \\3", "absdfds0213451"), " ")
[[1]]
[1] "absdfds" "0213451"

> strsplit(gsub("([a-z]*)([0-9]*)([a-z]*)", "\\1 \\2 \\3", "0213451ab"), " ")
[[1]]
[1] ""        "0213451" "ab"  


Using the 3 back references in the regex above will limit the parsing to up to 3 possible components. If you may have more than 3 you can increase the back reference sequence to some maximum number. However that can get tedious, so you may want to consider multiple passes using strsplit() to extract letters during one pass and then numbers during a second, or write a function to encapsulate that process.

Here are examples using strsplit():

# Get the numbers, using letters as the split
> strsplit("absdfds0213451ab", split = "[a-z]+")
[[1]]
[1] ""        "0213451"

> strsplit("absdfds0213451ab4567", split = "[a-z]+")
[[1]]
[1] ""        "0213451" "4567"   


# Get the letters, using numbers as the split
> strsplit("absdfds0213451ab", split = "[0-9]+")
[[1]]
[1] "absdfds" "ab"     

> strsplit("0213451ab", split = "[0-9]+")
[[1]]
[1] ""   "ab"

> strsplit("0213451ab123xyz789lmn", split = "[0-9]+")
[[1]]
[1] ""    "ab"  "xyz" "lmn"


Regards,

Marc


> 
> 
>> Regards,
>> 
>> Carol
>> 
>> 
>> On Wednesday, July 30, 2014 10:52 PM, Marc Schwartz <marc_schwartz at me.com> wrote:
>> 
>> 
>> 
>> On Jul 30, 2014, at 3:13 PM, carol white <wht_crl at yahoo.com> wrote:
>> 
>>> Hi,
>>> If I have a string of consecutive chars followed by consecutive numbers and then chars, like "absdfds0213451ab", how to separate the consecutive chars from consecutive numbers?
>>> 
>>> grep doesn't seem to be helpful
>>> 
>>> grep("[a-z]","absdfds0213451ab", ignore.case=T)
>>> [1] 1
>>> 
>>> 
>>>   grep("[0-9]","absdfds0213451ab", ignore.case=T)
>>> [1] 1
>>> 
>>> Thanks
>>> 
>>> Carol
>> 
>> 
>> grep() will only tell you that a pattern is present. You want to use gsub() or similar with back references to return parts of the vector.
>> 
>> Will they ALWAYS appear in that pattern (letters, numbers, letters) or is there some level of variation?
>> 
>> If they will always appear as in your example, then one approach is:
>> 
>>> strsplit(gsub("([a-z]+)([0-9]+)([a-z]+)", "\\1 \\2 \\3", "absdfds0213451ab"), " ")
>> 
>> [[1]]
>> [1] "absdfds" "0213451" "ab"
>> 
>> 
>> The initial gsub() returns the 3 parts separated by a space, which is then used as the split argument to strsplit().
>> 
>> If there will be some variation, you can use multiple calls to gsub() or similar, each getting either the letters or the numbers.
>> 
>> Regards,
>> 
>> Marc Schwartz