You can return a Series from the applied function that contains the new data, preventing the need to iterate three times. Passing axis=1 to the apply function applies the function sizes to each row of the dataframe, returning a series to add to a new dataframe. This series, s, contains the new values, as well … Read more
You can use applymap() which is concise for your case. df.applymap(foo_bar) # A B C #0 wow bar wow bar #1 bar wow wow bar Another option is to vectorize your function and then use apply method: import numpy as np df.apply(np.vectorize(foo_bar)) # A B C #0 wow bar wow bar #1 bar wow wow … Read more
This already exists: library(zoo) na.locf(data.frame(rep(“a”,4), 1:4,1:4, c(1,NA,NA,NA)))
following should do the trick cbind(id = x[, 1], x[, -1]/rowSums(x[, -1])) ## id val0 val1 val2 ## 1 a 0.08333333 0.3333333 0.5833333 ## 2 b 0.13333333 0.3333333 0.5333333 ## 3 c 0.16666667 0.3333333 0.5000000
The culprit is here. With func=tuple it works, but using func=list raises an exception from within the compiled module lib.reduce: ValueError: (‘function does not reduce’, 0) As you can see, they catch the exception but don’t bother to handle it. Even without the too-broad except clause, that’s a bug in pandas. You might try to … Read more
One option is ifelse which is vectorized version of if/else. If we are doing this for each row, the if/else as showed in the OP’s pandas post can be done in either a for loop or lapply/sapply, but that would be inefficient in R. df <- transform(df, c= ifelse(a==b, a+b, b-a)) df # a b … Read more
We can use ls with mget library(data.table) rbindlist(mget(ls(pattern = “^df\\.\\d+”))) Or with dplyr library(dplyr) mget(ls(pattern=”^df\\.\\d+”)) %>% bind_rows() Or with rbind from base R do.call(rbind, mget(ls(pattern=”^df\\.\\d+”)))
This may be more generalizable to your situation in that you pass a list of indices. If speed is an issue (large data frame) I’d opt for lapply with do.call rather than sapply: x <- list(1:3, 4:6) do.call(cbind, lapply(x, function(i) rowMeans(dat[, i]))) Works if you just have col names too: x <- list(c(‘a’,’b’,’c’), c(‘d’, ‘e’, … Read more
apply converts your data.frame to a character matrix. Use lapply: lapply(a, class) # $x1 # [1] “numeric” # $x2 # [1] “factor” # $x3 # [1] “factor” In second command apply converts result to character matrix, using lapply: a2 <- lapply(a, as.factor) lapply(a2, class) # $x1 # [1] “factor” # $x2 # [1] “factor” # … Read more
unname(tapply(v, (seq_along(v)-1) %/% n, sum)) # [1] 55 155 255 355 455 555 655 755 855 955