generating 1000 names with replacement using r_data_frame about wakefield HOT 8 CLOSED

To use arguments to a variable function you need to use parenthesis as in:

x <- r_data_frame(n =1000,
    name(replace = TRUE)
)

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I'm closing this feel free to comment. It would help future posters if you let them know if this solved your issue.

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Yes, this solved the single variable issue. Thank you.
However multiple variables do no work with this syntax. Maybe there is a workaround.

x <- data.frame(
name(10000, replace = TRUE),
sex(10000, replace = TRUE)
)

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Why do you need replace = TRUE on the sex variable? Also why are you using data.frame?

from wakefield.

started with r_data_frame, but I need sampling with replacement. Name per se not needed, but thats the variable in which I originally encountered the problem. I can work around it, it's just not knowing what parameters to use to consistently cast it to a data frame.

#works with N < 1000
r_data_frame(
n = 100,
name,
sex
)

from wakefield.

I don't understand. What I showed above works as you want I believe... just add new variables to it as follows:

r_data_frame(
    n = 10000,
    name(replace = TRUE),
    sex
)

from wakefield.

OK. That works.
But, then switching back to no replace, how would you could for error in?:

y <- r_data_frame(

• n = 10000,
• name(replace = FALSE),
• sex
• )
  Error in sample.int(length(x), size, replace, prob) :
  cannot take a sample larger than the population when 'replace = FALSE'

#Works for n<1000

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There are only about 1000 names in the vector I select from. If you don't replace and want a sample larger than this it will throw an error. You have two options

1. use replacement
2. create your own names function with a vector that contains more names

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