How can I replace a specific value (e.g. -10000) with NA within a big list?

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This works for the first list element:

values[[1]][values[[1]]==-10000] <-NA

As I do not want to loop over my thousand list elements I'm looking for a command which does the same for the whole list like:

 values[values==-10000] <-NA

But this does not work for the type list:

Error in values[values == -10000] <- NA : (list) object cannot be coerced to type 'double'

Try using lapply:

lst <- list(v1=-10000, v2=500, v3=c(1,2))
lapply(lst, function(x) ifelse(x==-10000, NA, x))

 [1] NA

[1] 500

[1] 1 2

This approach is also robust even if some of the list elements are not numbers, but are other things, such as vectors. In that case, a vector would not match to your target value, and would not be changed.

pandas.Series.replace, Replace values given in to_replace with value . Values in place. Note: this will modify any other views on this object (e.g. a column from a DataFrame). If to_replace is a dict and value is not a list , dict , ndarray , or Series Fill NA values. to_replace : [str, regex, list, dict, Series, numeric, or None] pattern that we are trying to replace in dataframe. value : Value to use to fill holes (e.g. 0), alternately a dict of values specifying which value to use for each column (columns not in the dict will not be filled). Regular expressions, strings and lists or dicts of such objects

As you mention "big list" I provide a second option that uses replace instead which is a bit quicker compared to ifelse.

Thanks to @TimBiegeleisen for the data

lst <- list(v1=-10000, v2=500, v3=c(1,2))
lapply(lst, function(x) replace(x, x == -10000, NA))
#[1] NA
#[1] 500
#[1] 1 2


l <- rep(lst, 100000)
benchmark <- microbenchmark(
  tim = lapply(l, function(x) ifelse(x==-10000, NA, x)),
  markus = lapply(l, function(x) replace(x, x==-10000, NA))


#Unit: milliseconds
#   expr      min        lq      mean    median        uq      max neval cld
#    tim 931.5551 1003.0364 1054.7647 1018.7956 1082.3210 2536.373   100   b
# markus 432.3821  473.9881  500.4833  482.5838  515.9907 1023.392   100  a 

Replace all 0 values to NA, NA is a logical constant of length 1 which contains a missing value indicator. Now you could replace zeroes by NULL in a data frame in the sense of completely When using, e.g., var , cov , or cor , that is actually equivalent to first replacing Table version of this, and because the given data.frame solution does not work  I can create 1 dependant list using the INDIRECT formula without problem but I cannot get any additional lists created that are dependent on the first selection due to not being able to define the names in the same way, i.e. select UK as country then the next list is defined by UK but I cannot create a third due to not being able to name

Using na_if with map

map(lst, na_if, y = -10000)
#[1] NA

#[1] 500

#[1] 1 2
lst <- list(v1=-10000, v2=500, v3=c(1,2))

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R Find Missing Values (6 Examples for Data Frame, Column & Vector), How to find missing data in R – Identify NA values in vectors, data frames or matrices Example 2: Find missing values in a column of a data frame columns by squared brackets which([ , 2])) # Beside the change from might arise how to count missing values per row, by column, or in a single vector. The dplyr hybridized options are now around 30% faster than the Base R subset reassigns. On a 100M datapoint dataframe mutate_all(~replace(.,, 0)) runs a half a second faster than the base R d[] <- 0 option.

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