将函数应用于列表中所有数据框的某些列，然后将值分配给列 [英] apply function to certain columns of all dataframe in list and then assign value to columns

问题描述

在此处

我有一个数据帧列表(即1000个)，如下所示:

I have a list of data frames (i.e. 1000) that looks like this:

> lst[1000]
$rand1000
                        Binomial         S4         S5    S6
254            Euastacus suttoni  25.816165  4.1916527  86.3
327            Orconectes hobbsi  16.726409  5.5241308  25.0
282           Faxonella creaseri  17.026970  6.4109494  18.0
319           Orconectes durelli  14.093957  7.2342324  35.0
525           Procambarus raneyi  15.799001  6.3746574  55.1

我想对列表中的所有数据帧的S4，S5和S6列应用功能.所以我写了这个函数:

I want to apply function to S4, S5 and S6 columns for all data frames in the list. So I wrote this function:

lapply(lst, function(x) {x$S4 <- sensitivity.rand(x[[2]], 25); x})

(此处的"sensitivity.rand"是采用 vector 和 threshold (在上面的代码25中是阈值)并分配为H或L的函数)

(here 'sensitivity.rand' is a function that takes the vector and threshold (in the above code 25 is the threshold) and assign into H or L)

给出的输出为(很好):

that give the output as (which is good):

$rand1000
                        Binomial S4         S5    S6
254            Euastacus suttoni  H  4.1916527  86.3
327            Orconectes hobbsi  H  5.5241308  25.0
282           Faxonella creaseri  H  6.4109494  18.0
319           Orconectes durelli  H  7.2342324  35.0
525           Procambarus raneyi  H  6.3746574  55.1

但是，如果我看到原始数据帧没有更改(即，它像以前一样).我怎样才能做到这一点.我需要对列表中所有数据框的所有S4，S5和S6列执行此操作.这样原始数据帧将变成这样:

But if I see that the original data frame is not changed (i.e. it is as it was before). How can I do this. I need to do this for all S4, S5 and S6 columns for all data frame in the list. So that the original data frames will change into like this:

> lst[1000]
$rand1000
                            Binomial S4  S5 S6
    254            Euastacus suttoni  H  H  H
    327            Orconectes hobbsi  H  L  H
    282           Faxonella creaseri  H  H  L
    319           Orconectes durelli  H  L  L
    525           Procambarus raneyi  H  H  H

推荐答案

我们需要将输出分配回 list 或创建一个新对象.同样，在OP的代码中，该功能仅适用于第二列.我们可以遍历感兴趣的列并应用该函数，也可以单独执行

We need to assign the output back to the list or create a new object. Also, in the OP's code, the function is only applied to the 2nd column. We can either loop over the columns of interest and apply the function or do it separately

lst <- lapply(lst, function(x) {x[2:4] <- lapply(x[2:4], sensitivity.rand, threshold = 25)
                 x})

---

或者如果我们使用的是 tidyverse ，则可以使用 mutate_each

lst <- lapply(lst, function(x) x %>%
                                 mutate_each(funs(sensitivity.rand(., 25)), 2:4))

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