我意识到此前已经提出过这样的问题，但我不明白为什么我的代码会破坏。
我已尝试mapply单独使用do.call以及purrr包的pmap功能。我不断得到“未使用的参数”错误。由于所有3个都在失败，我想我必须在参数中错误地引用我的数据。我使用mdply包中的plyr做了类似的事情，但那是一年多以前的事了。当然，任何替代方法也会受到赞赏。
要创建数据框，compar：
obs = floor(runif(500, 1,99))
p = round(runif(500,0,1), digits = 4)
n = floor(runif(500, 100,150))
test = rep("two.sided", 500)
conf = rep(0.95, 500)
compar = as.data.frame(cbind(obs,n, p))
compar$test = test
compar$conf = conf
head(compar, 3)
obs p n test conf
1 47 0.2432 133 two.sided 0.95
2 52 0.3391 118 two.sided 0.95
3 22 0.2790 115 two.sided 0.95
我尝试pmap：
pmap(.l = compar, .f = binom.test)
Error in .f(obs = .l[[c(1L, i)]], p = .l[[c(2L, i)]], n = .l[[c(3L, i)]], :
unused arguments (obs = .l[[c(1, i)]], test = .l[[c(4, i)]])
接下来，mapply：
mapply(compar, FUN = binom.test)
Error in (function (x, n, p = 0.5, alternative = c("two.sided", "less", :
incorrect length of 'x'
最后，do.call和mapply
do.call(mapply, c(binom.test, compar[c("obs", "n", "p", "test", "conf")]))
Error in (function (x, n, p = 0.5, alternative = c("two.sided", "less", :
unused arguments (obs = dots[[1]][[1]], test = dots[[4]][[1]])
答案 0 :(得分：1)
列名与binom.test个参数不匹配;对于pmap版本，根据binom.test参数重命名列应该有效：
pmap(select(compar, x=obs, n, p, alternative=test, conf), binom.test)
#[[1]]
# Exact binomial test
#data: .l[[c(1L, i)]] and .l[[c(2L, i)]]
#number of successes = 5, number of trials = 149, p-value < 2.2e-16
#alternative hypothesis: true probability of success is not equal to 0.435
#95 percent confidence interval:
# 0.01098400 0.07657136
#sample estimates:
#probability of success
# 0.03355705
#[[2]]
# Exact binomial test
#data: .l[[c(1L, i)]] and .l[[c(2L, i)]]
#number of successes = 20, number of trials = 113, p-value = 1.391e-10
#alternative hypothesis: true probability of success is not equal to 0.4681
#95 percent confidence interval:
# 0.1115928 0.2600272
#sample estimates:
#probability of success
# 0.1769912
# more output
或者：pmap(rename(compar, x=obs, alternative=test), binom.test)