I try to replace the german special character "ö" in a dataframe by "oe". The charcter occurs in multiple columns so I would like to be able to do this all in one by not having to specify individual columns.
Here is a small example of the data frame
data <- data.frame(a=c("aö","ab","ac"),b=c("bö","bb","ab"),c=c("öc","öb","acö"))
I tried :
data[data=="ö"]<-"oe"
but this did not work since I would need to work with regular expressions here. However when I try :
data[grepl("ö",data)]<-"oe"
I do not get what I want.
The dataframe at the end should look like:
> data
a b c
1 aoe boe oec
2 ab bb oeb
3 ac ab acoe
>
The file is a csv import that I import by read.csv. However, there seems to be no option to change to fix this with the import statement.
How do I get the desired outcome?
Here's one way to do it:
data <- apply(data,2,function(x) gsub("ö",'oe',x))
Explanation:
Your grepl doesn't work because grepl just returns a boolean matrix (TRUE/FALSE) corresponding to the elements in your data frame for which the regex matches. What the assignment then does is replace not just the character you want replaced but the entire string. To replace part of a string, you need sub (if you want to replace just once in each string) or gsub (if you want all occurrences replaces). To apply that to every column you loop over the columns using apply.
If you want to return a data frame, you can use:
data.frame(lapply(data, gsub, pattern = "ö", replacement = "oe"))
Related
I have a very large pandas data frame containing both string and integer columns. I'd like to search the whole data frame for a specific substring, and if found, replace the full string with something else.
I've found some examples that do this by specifying the column(s) to search, like this:
df = pd.DataFrame([[1,'A'], [2,'(B,D,E)'], [3,'C']],columns=['Question','Answer'])
df.loc[df['Answer'].str.contains(','), 'Answer'] = 'X'
But because my data frame has dozens of string columns in no particular order, I don't want to specify them all. As far as I can tell using df.replace will not work since I'm only searching for a substring. Thanks for your help!
You can use data frame replace method with regex=True, and use .*,.* to match strings that contain a comma (you can replace comma with other any other substring you want to detect):
str_cols = ['Answer'] # specify columns you want to replace
df[str_cols] = df[str_cols].replace('.*,.*', 'X', regex=True)
df
#Question Answer
#0 1 A
#1 2 X
#2 3 C
or if you want to replace all string columns:
str_cols = df.select_dtypes(['object']).columns
I have a file of 1000 lines, with 5 to 8 columns in each line separated by :
1:2:3:4:5:6:7:8
4g10:8s:45:9u5b:a:z1
I want to have all lines in some order 4:3:1:2:5:6:7...
How would I swap only first 4 columns with regex?
I think this would probably be easier to do with another approach, but you could use ex to do it, so be in command mode and enter:
:%s/^\([^:]\+\):\([^:]\+\):\([^:]\+\):\([^:]\+\):/\4:\3:\1:\2:/
which will create capture groups for the first 4 colon delimited fields, then replace them in a different order than they were there originally.
Here is a regex that should do what you are looking for:
newtext = re.sub("([^:]+):([^:]+):([^:]+):([^:]+)(:)?(.*)?",r"\4:\3:\1:\2\5\6",text)
The take away is you'll want to use parans for capturing and then reorder them in the order you want them in the replace. Each capture "group" is just one or more non : separated by : If there is possibility of empty groups change each + to a *
Here is a sample in Python for clarity:
import re
textlist = [
"1:2:3:4:5:6:7:8",
"1:2:3:4:5",
"1:2:3:4",
]
for text in textlist:
newtext = re.sub("([^:]+):([^:]+):([^:]+):([^:]+)(:)?(.*)?",r"\4:\3:\1:\2\5\6",text)
print (newtext)
output:
4:3:1:2:5:6:7:8
4:3:1:2:5
4:3:1:2
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
General situation: I am currently trying to name dataframes inside a list in accordance to the csv files they have been retrieved from, I found that using gsub and regex is the way to go. Unfortunately, I can’t produce exactly what I need, just sort of.
I would be very grateful for some hints from someone more experienced, maybe there is a reasonable R regex cheat cheet ?
File are named r2_m1_enzyme.csv, the script should use the first 4 characters to name the corresponding dataframe r2_m1, and so on…
# generates a list of dataframes, to mimic a lapply(f,read.csv) output:
data <- list(data.frame(c(1,2)),data.frame(c(1,2)),data.frame(c(1,2)),data.frame(c(1,2)))
# this mimics file names obtained by list.files() function
f <-c("r1_m1_enzyme.csv","r2_m1_enzyme.csv","r1_m2_enzyme.csv","r2_m2_enzyme.csv")
# this should name the data frames according to the csv file they have been derived from
names(data) <- gsub("r*_m*_.*","\\1", f)
but it doesnt work as expected... they are named r2_m1_enzyme.csv instead of the desired r2_m1, although .* should stop it?
If I do:
names(data) <- gsub("r*_.*","\\1", f)
I do get r1, r2, r3 ... but I am missing my second index.
The question: So my questions is, what regex expression would allow me to obtain strings “r1_m1”, “r2_m1”, “r1_m2”, ... from strings that are are named r*_m*_xyz.csv
Search history: R regex use * for only one character, Gsub regex replacement, R ussing parts of filename to name dataframe, R regex cheat sheet,...
If your names are always five characters long you could use substr:
substr(f, 1, 5)
If you want to use gsub you have to group your expression (via ( and )) because \\1 refers to the first group and insert its content, e.g.:
gsub("^(r[0-9]+_m[0-9]+).*", "\\1", f)
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/