英文:
R generate random data from several distributions using kwargs
问题
问题：
我想从多个分布中生成随机样本，例如norm、pois等。然而，我想要指定给rdist函数的参数，就像传递参数给函数调用一样。
尝试：
我尝试了一些我在Python中做的事情，例如partial或**kwargs
dist = 'rnorm' # 分布n = 100 # 样本大小para1 = c(0,1) # 参数mean(get(dist)(n,para1)) # 失败para2 = c(mean = 0, sd = 1) # 参数mean(get(dist)(n,para2)) # 失败
输出:
> ```R
[1] 0.6318898
[2] 0.4246129
从输出中可以看出，rnorm没有生成来自norm(mean = 0,sd = 1)的样本，因为样本均值约为0.5。输出的随机数是分别从rnorm(1,mean = 0,sd = 1)和rnorm(1,mean = 1, sd =1)生成的。
英文:
Question:
I want to generate random sample from several distributions, e.g. norm, pois etc. However, I want to specify parameters to rdist function, like passing parameters to function call.
Attempt:
I tried with something I did in python partial or **kwargs
dist = 'rnorm' # distn = 100 # sample sizepara1 = c(0,1) # parametersmean(get(dist)(n,para1)) # it failspara2 = c(mean = 0, sd = 1)# parametersmean(get(dist)(n,para2)) # it fails
Output:
> ```
[1] 0.6318898
[2] 0.4246129
As we can see from the output, rnorm did not produce samples from norm(mean = 0,sd = 1),beacause sample mean is around 0.5. The output randoms are generated from rnorm(1,mean = 0,sd = 1) and rnorm(1,mean = 1, sd =1) one by one.
答案1
得分: 1
你可以使用 do.call 来调用一个函数作为字符名称，并将函数参数传递在一个列表中（还可参考 ?do.call）：
## 从标准正态分布中抽样(X <- do.call("rnorm", args = list(n = 10, mean = 0, sd = 1)))#> [1] 1.4127136 -0.4113396 1.7546034 0.6741983 -1.3499156 -0.8033819#> [7] -1.6393473 -0.7667389 0.7157213 -1.1061452mean(X)#> [1] -0.1519632
英文:
You can use do.call to call a function as a character name and pass function arguments in a list (see also ?do.call):
## sample from standard normal(X <- do.call("rnorm", args = list(n = 10, mean = 0, sd = 1)))#> [1] 1.4127136 -0.4113396 1.7546034 0.6741983 -1.3499156 -0.8033819#> [7] -1.6393473 -0.7667389 0.7157213 -1.1061452mean(X)#> [1] -0.1519632
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