I would like to conditionally format the first instance in a column where the value matches the criteria (in the example below >= 1000000).
Here is an example spreadsheet:
https://docs.google.com/spreadsheets/d/1Bz2jHn_rQneb62QLuSGv46PdOv7V5SennmeL3vQcPbo/edit#gid=0
try:
=(INDEX(COUNTIFS(IF(A$2:A>=1000000, 1), IF(A2>=1000000, 1),
ROW(A$2:A), "<="&ROW(A2)))=1)*(A2>=1000000)
Related
So in my PostgreSQL 10 I have a column of type integer. This column represents a code of products and it should be searched against another code or part of the code. The values of the column are made of three parts, a five-digit part and two two-digit parts. Users can search for only the first part, the first-second or first-second-third.
So, in my column I have , say 123451233 the user searches for 12345 (the first part). I want to be able to return the 123451233. Same goes if the users also searches for 1234512 or 123451233.
Unfortunately I cannot change the type of column or break the one column into three (one for every part). How can I do this? I cannot use LIKE. Maybe something like a regex for integers?
Thanks
Consider to use simple arithmetic.
log(value)::int + 1 returns the number of digits in integer part of the value and using this:
value/(10^(log(value)::int-log(search_input)::int))::int
returns value truncated to the same digits number as search_input so, finally
search_input = value/(10^(log(value)::int-log(search_input)::int))::int
will make the trick.
It is more complex literally but also could be more efficient then strings manipulations.
PS: But having index like create index idx on your_table(cast(your_column as text)); search like
select * from your_table
where cast(your_column as text) like search_input || '%';
is the best case IMO.
You do not need regex functions. Cast the integer to text and use the function left(), example:
create table my_table(code int); -- or bigint
insert into my_table values (123451233);
with input_data(input_code) as (
values('1234512')
)
select t.*
from my_table t
cross join input_data
where left(code::text, length(input_code)) = input_code;
code
-----------
123451233
(1 row)
I have several tables that I'm trying to update. I need to update the fourth element of each table, when the first element matches a certain string.
Example table (in Lua):
TABLE = {
{TEXT1,0,4,0,0,0,0},
{TEXT2,0,4,0,0,0,0},
{TEXT3,0,5,0,0,0,0},
{TEXT1,0,5,0,0,0,0},
{TEXT1,0,4,0,0,0,0},
{TEXT4,0,5,0,0,0,0}
}
I can use the following to select only tables that have the element "TEXT1":
(?=TEXT1)
I can also use the following statement to select the content in between the fourth and fifth comma:
^[^,]*,[^,]*,[^,]*,[^,]*,\K(.*?)(?=,\d)
I can't figure out how to use both expressions at once though.
So basically, I'd like to replace 0 in every table that starts with the string "TEXT1", but how do I combine my statements to check for the required text, then select the fourth element in that table only?
You can use ^\{TEXT1,([^,]+,){3}\K[^,]+ to select the element you are after and then replace it by whatever you like. Replacing by 999 would turn your table into
TABLE = {
{TEXT1,0,4,0,999,0,0},
{TEXT2,0,4,0,0,0,0},
{TEXT3,0,5,0,0,0,0},
{TEXT1,0,5,0,999,0,0},
{TEXT1,0,4,0,999,0,0},
{TEXT4,0,5,0,0,0,0}
}
Note that in your title ask for the fourth element, but your regex selects the element betweeen fourth and fifth comma - which is the fifth element and used here. Also note that you might want to replace [^,] by [^,\n\r] if the table could contain less than six elements at any place.
I need to create some columns from a cell that contains text separated by "_".
The input would be:
campaign1_attribute1_whatever_yes_123421
And the output has to be in different columns (one per field), with no "_" and excluding the final number, as it follows:
campaign1 attribute1 whatever yes
It must be done using a regex formula!
help!
Thanks in advance (and sorry for my english)
=REGEXEXTRACT("campaign1_attribute1_whatever_yes_123421","(("®EXREPLACE("campaign1_attribute1_whatever_yes_123421","((_)|(\d+$))",")$1(")&"))")
What this does is replace all the _ with parenthesis to create capture groups, while also excluding the digit string at the end, then surround the whole string with parenthesis.
We then use regex extract to actuall pull the pieces out, the groups automatically push them to their own cells/columns
To solve this you can use the SPLIT and REGEXREPLACE functions
Solution:
Text - A1 = "campaign1_attribute1_whatever_yes_123421"
Formula - A3 = =SPLIT(REGEXREPLACE(A1,"_+\d*$",""), "_", TRUE)
Explanation:
In cell A3 We use SPLIT(text, delimiter, [split_by_each]), the text in this case is formatted with regex =REGEXREPLACE(A1,"_+\d$","")* to remove 123421, witch will give you a column for each word delimited by ""
A1 = "campaign1_attribute1_whatever_yes_123421"
A2 = "=REGEXREPLACE(A1,"_+\d*$","")" //This gives you : *campaign1_attribute1_whatever_yes*
A3 = SPLIT(A2, "_", TRUE) //This gives you: campaign1 attribute1 whatever yes, each in a separate column.
I finally figured it out yesterday in stackoverflow (spanish): https://es.stackoverflow.com/questions/55362/c%C3%B3mo-separo-texto-por-guiones-bajos-de-una-celda-en...
It was simple enough after all...
The reason I asked to be only in regex and for google sheets was because I need to use it in Google data studio (same regex functions than spreadsheets)
To get each column just use this regex extract function:
1st column: REGEXP_EXTRACT(Campaña, '^(?:[^_]*_){0}([^_]*)_')
2nd column: REGEXP_EXTRACT(Campaña, '^(?:[^_]*_){1}([^_]*)_')
3rd column: REGEXP_EXTRACT(Campaña, '^(?:[^_]*_){2}([^_]*)_')
etc...
The only thing that has to be changed in the formula to switch columns is the numer inside {}, (column number - 1).
If you do not have the final number, just don't put the last "_".
Lastly, remember to do all the calculated fields again, because (for example) it gets an error with CPC, CTR and other Adwords metrics that are calculated automatically.
Hope it helps!
I have a large dataframe (3000+ columns) and I am trying to get a list of all column names that follow this pattern:
"stat.mineBlock.minecraft.123456stone"
"stat.mineBlock.minecraft.DFHFFBSBstone2"
"stat.mineBlock.minecraft.AAAstoneAAAA"
My code:
stoneCombined<-grep("^[stat.mineBlock.minecraft.][a-zA-Z0-9]*?[stone][a-zA-Z0-9]*?", colnames(ingame), ignore.case =T)
where ingame is the dataframe I am searching. My code returns a list of numbers however instead of the dataframe columns (like those above) that I was expecting. Con someone tell me why?
After adding value=TRUE (Thanks to user227710):
I now get column names, but I get every column in my dataset not those that contain : stat.mineBlock.minecraft. and stone like I was trying to get.
To return the column names you need to set value=TRUE as an additional argument of grep. The default option in grep is to set value=FALSE and so it will give you indices of the matched colnames. .
help("grep")
value
if FALSE, a vector containing the (integer) indices of the matches determined by grep is returned, and if TRUE, a vector containing the matching elements themselves is returned.
grep("your regex pattern", colnames(ingame),value=TRUE, ignore.case =T)
Here is a solution in dplyr:
library(dplyr)
your_df %>%
select(starts_with("stat.mineBlock.minecraft"))
The more general way to match a column name to a regex is with matches() inside select(). See ?select for more information.
My answer is based on this SO post. As per the regex, you were very close.
Just [] create a character class matching a single character from the defined set, and it is the main reason it was not working. Also, perl=T is always safer to use with regex in R.
So, here is my sample code:
df <- data.frame(
"stat.mineBlock.minecraft.123456stone" = 1,
"stat.mineBlock.minecraft.DFHFFBSBwater2" = 2,
"stat.mineBlock.minecraft.DFHFFBSBwater3" = 3,
"stat.mineBlock.minecraft.DFHFFBSBstone4" = 4
)
grep("^stat\\.mineBlock\\.minecraft\\.[a-zA-Z0-9]*?stone[a-zA-Z0-9]*?", colnames(df), value=TRUE, ignore.case=T, perl=T)
See IDEONE demo
I want to apply a search and replace regular expression pattern that work only in a given range of line and column on a text file like this :
AAABBBFFFFBBBAAABBB
AAABBBFFFFBBBAAABBB
GGGBBBFFFFBHHAAABBB
For example i want to replace BBB with YYY in line range 1 to 2 and from column 4 to 6, then obtaining this output :
AAAYYYFFFFBBBAAABBB
AAAYYYFFFFBBBAAABBB
GGGBBBFFFFBHHAAABBB
Is there a way to do it with Vim ?
:1,2 s/\%3cBBB/YYY/
\%3c means third column (see :help /\%c or more globally :help pattern)
If this is always the first one you want to replace, simply don't specify /g
:1,2s/BBB/YYY/
would work fine.
Alternatively, if you need to exactly specify which column you want replaced, you can use the \%Nv syntax, where N is the virtual column (column as it looks, so tabs are multiple columns, use c instead of v for actual columns)
Replacing the second set of B's on lines 1 and 2 could be done with:
:1,2s/\%11vBBB/YYY/