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每当我将 purrr 库函数pmap()与包含具有多个条件的 if 语句的函数一起使用时，if 语句似乎无法正常工作。为了向您展示我的意思，这是一个使用 gapminder 数据集的可重现示例。

library(tidyverse)
library(gapminder)
library(broom)

# Nest the tibble into separate dataframes for each country-continent combination

by_country <- gapminder %>%
  group_by(country, continent) %>%
  nest()

现在我想为每个分组数据框建立一个线性回归模型。问题是我想根据国家和大陆在我的模型中使用不同的 x 变量。这是我的函数，我怀疑 if 语句有问题：

# My function
country_model <- function(df, cont, count) {

  if(cont == "Asia" & count == "Afghanistan") { # 2 conditions

    lm(lifeExp ~ year, data = df)

  } else {

    lm(lifeExp ~ pop, data = df)

  }
}

现在我将采用该功能并将其应用于所有分组的数据帧。我期望模型汇总输出将显示阿富汗数据集的模型将有一个系数year而不是pop.

by_country2 <- by_country %>%
  mutate(model = pmap(list(data, continent, country), country_model),
         modelsum = map(model, tidy)) %>%
  unnest(modelsum, .drop = TRUE)

by_country2

我的输出显示阿富汗的系数是pop，不是year。

A tibble: 284 × 7
       country continent        term     estimate    std.error  statistic      p.value
        <fctr>    <fctr>       <chr>        <dbl>        <dbl>      <dbl>        <dbl>
1  Afghanistan      Asia (Intercept) 2.834615e+01 2.314395e+00  12.247758 2.410050e-07
2  Afghanistan      Asia         pop 5.771517e-07 1.343425e-07   4.296121 1.570999e-03
3      Albania    Europe (Intercept) 4.963274e+01 1.935933e+00  25.637630 1.871817e-10
4      Albania    Europe         pop 7.286188e-06 7.171585e-07  10.159802 1.374311e-06
5      Algeria    Africa (Intercept) 3.565187e+01 1.632853e+00  21.834099 9.087006e-10
6      Algeria    Africa         pop 1.176242e-06 7.588190e-08  15.500960 2.548769e-08
7       Angola    Africa (Intercept) 2.855043e+01 1.922225e+00  14.852803 3.843692e-08
8       Angola    Africa         pop 1.276860e-06 2.482137e-07   5.144195 4.351004e-04
9    Argentina  Americas (Intercept) 5.323586e+01 3.784907e-01 140.653008 8.102227e-18
10   Argentina  Americas         pop 5.532629e-07 1.282987e-08  43.123018 1.079775e-12
# ... with 274 more rows

对我来说奇怪的是，当我在函数 if 语句中只使用 1 个条件时，它似乎工作得很好：

country_model <- function(df, cont) {

  if(cont == "Asia") { # Only 1 condition

    lm(lifeExp ~ year, data = df)

  } else {

    lm(lifeExp ~ pop, data = df)

  }
}

by_country2 <- by_country %>%
  mutate(model = map2(data, continent, country_model),
         modelsum = map(model, tidy)) %>%
  unnest(modelsum, .drop = TRUE)

by_country2

# A tibble: 284 × 7
       country continent        term      estimate    std.error  statistic      p.value
        <fctr>    <fctr>       <chr>         <dbl>        <dbl>      <dbl>        <dbl>
1  Afghanistan      Asia (Intercept) -5.075343e+02 4.048416e+01 -12.536613 1.934055e-07
2  Afghanistan      Asia        year  2.753287e-01 2.045093e-02  13.462890 9.835213e-08
3      Albania    Europe (Intercept)  4.963274e+01 1.935933e+00  25.637630 1.871817e-10
4      Albania    Europe         pop  7.286188e-06 7.171585e-07  10.159802 1.374311e-06
5      Algeria    Africa (Intercept)  3.565187e+01 1.632853e+00  21.834099 9.087006e-10
6      Algeria    Africa         pop  1.176242e-06 7.588190e-08  15.500960 2.548769e-08
7       Angola    Africa (Intercept)  2.855043e+01 1.922225e+00  14.852803 3.843692e-08
8       Angola    Africa         pop  1.276860e-06 2.482137e-07   5.144195 4.351004e-04
9    Argentina  Americas (Intercept)  5.323586e+01 3.784907e-01 140.653008 8.102227e-18
10   Argentina  Americas         pop  5.532629e-07 1.282987e-08  43.123018 1.079775e-12
# ... with 274 more rows

我不确定我的问题是与pmap()我的 if 语句有关。

4

1 回答 1

1

这与此 GitHub 问题有关。

它似乎pmap是通过数字发送的continent，country这可以通过在您的函数中放置一个打印语句来确认。

test_fun <- function(df, cont, xx) {
  print(paste(cont, xx))
}

temp <-by_country %>%
  mutate(model = pmap(list(data, continent, country), test_fun))

印刷：

[1] "3 1"
[1] "4 2"
[1] "1 3"
[1] "1 4"
[1] "2 5"
[1] "5 6"
[1] "4 7"
[1] "3 8"
[1] "3 9"
etc

这不会发生在 中map2，因此您的第二次尝试确实有效。

强制字符解决了这个问题：

by_country %>%
  mutate(model = pmap(list(data, as.character(continent), as.character(country)), country_model),
         modelsum = map(model, broom::tidy)) %>%
  unnest(modelsum, .drop = TRUE)

# A tibble: 284 x 7
       country continent        term      estimate    std.error  statistic      p.value
        <fctr>    <fctr>       <chr>         <dbl>        <dbl>      <dbl>        <dbl>
 1 Afghanistan      Asia (Intercept) -5.075343e+02 4.048416e+01 -12.536613 1.934055e-07
 2 Afghanistan      Asia        year  2.753287e-01 2.045093e-02  13.462890 9.835213e-08
 3     Albania    Europe (Intercept)  4.963274e+01 1.935933e+00  25.637630 1.871817e-10
 4     Albania    Europe         pop  7.286188e-06 7.171585e-07  10.159802 1.374311e-06
 5     Algeria    Africa (Intercept)  3.565187e+01 1.632853e+00  21.834099 9.087006e-10
 6     Algeria    Africa         pop  1.176242e-06 7.588190e-08  15.500960 2.548769e-08
 7      Angola    Africa (Intercept)  2.855043e+01 1.922225e+00  14.852803 3.843692e-08
 8      Angola    Africa         pop  1.276860e-06 2.482137e-07   5.144195 4.351004e-04
 9   Argentina  Americas (Intercept)  5.323586e+01 3.784907e-01 140.653008 8.102227e-18
10   Argentina  Americas         pop  5.532629e-07 1.282987e-08  43.123018 1.079775e-12
# ... with 274 more rows
于 2017-05-24T11:29:05.150 回答