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r - 根据多个条件汇总计数

问题描述

我试图根据两个变量的组合来总结我的数据。以下代码用于处理数据：

df <- data_frame(fc = runif(1000, -5, 5),
           padj = runif(1000, 0, 1))

df %>% 
  summarise(
    dn_red = count(fc < -1.5, padj <= 0.1),
    dn_pink = count(fc < -1.5, padj >= 0.1),
    dn_blue = count(fc>-1.5 & fc< 0, padj <= 0.1),
    dn_grey = count(fc>-1.5 & fc< 0, padj >= 0.1),
    up_red = count(fc > 1.5, padj <= 0.1),
    up_pink = count(fc > 1.5, padj >= 0.1),
    up_blue = count(fc < 1.5 & fc > 0, padj <= 0.1),
    up_grey = count(fc < 1.5 & fc > 0, padj >= 0.1)
  )

在编写它几个月后运行它会引发以下错误：

Error: Problem with `summarise()` input `dn_red`.
x no applicable method for 'count' applied to an object of class "logical"
ℹ Input `dn_red` is `count(fc < -1.5, padj <= 0.1)`.

我可以看到 count 输出一个带有与条件相对应的逻辑向量的小标题。我试图从中得到的是计数的摘要，其中两个条件都为真。上面的代码曾经这样做......

标签： rtidyverse

解决方案

你也许想要sum代替count!

set.seed(1)
df <- data.frame(fc = runif(1000, -5, 5),
                 padj = runif(1000, 0, 1))

df %>% 
  summarise(
    dn_red = sum(fc < -1.5, padj <= 0.1),
    dn_pink = sum(fc < -1.5, padj >= 0.1),
    dn_blue = sum(fc>-1.5 & fc< 0, padj <= 0.1),
    dn_grey = sum(fc>-1.5 & fc< 0, padj >= 0.1),
    up_red = sum(fc > 1.5, padj <= 0.1),
    up_pink = sum(fc > 1.5, padj >= 0.1),
    up_blue = sum(fc < 1.5 & fc > 0, padj <= 0.1),
    up_grey = sum(fc < 1.5 & fc > 0, padj >= 0.1)
  )

  dn_red dn_pink dn_blue dn_grey up_red up_pink up_blue up_grey
1    494    1250     269    1025    458    1214     267    1023

但这会造成重叠。因此，您需要,在逻辑条件下替换为&或者|视情况而定。看。

df %>% 
  summarise(
    dn_red = sum(fc < -1.5 & padj <= 0.1),
    dn_pink = sum(fc < -1.5 & padj >= 0.1),
    dn_blue = sum(fc>-1.5 & fc< 0 & padj <= 0.1),
    dn_grey = sum(fc>-1.5 & fc< 0 & padj >= 0.1),
    up_red = sum(fc > 1.5 & padj <= 0.1),
    up_pink = sum(fc > 1.5 & padj >= 0.1),
    up_blue = sum(fc < 1.5 & fc > 0 & padj <= 0.1),
    up_grey = sum(fc < 1.5 & fc > 0 & padj >= 0.1)
  )

  dn_red dn_pink dn_blue dn_grey up_red up_pink up_blue up_grey
1     44     328      20     127     40     296      18     127

如果这是您所期望的，那么建议将1000数据点分成八种颜色。请改用此代码

df %>% mutate(new = case_when(
  fc < -1.5 & padj <= 0.1 ~ 'dn_red',
  fc < -1.5 & padj >= 0.1 ~ 'dn_pink',
  fc > -1.5 & fc < 0 & padj <= 0.1 ~ 'dn_blue',
  fc > -1.5 & fc < 0 & padj >= 0.1 ~'dn_grey',
  fc > 1.5 & padj <= 0.1 ~ 'up_red',
  fc > 1.5 & padj >= 0.1 ~ 'up_pink',
  fc < 1.5 & fc > 0 & padj <= 0.1 ~ 'up_blue',
  fc < 1.5 & fc > 0 & padj >= 0.1 ~ 'up_grey',
  TRUE ~ 'others'
)) %>% count(new)

      new   n
1 dn_blue  20
2 dn_grey 127
3 dn_pink 328
4  dn_red  44
5 up_blue  18
6 up_grey 127
7 up_pink 296
8  up_red  40

或更好地使用janitor频率计数

df %>% mutate(new = case_when(
  fc < -1.5 & padj <= 0.1 ~ 'dn_red',
  fc < -1.5 & padj >= 0.1 ~ 'dn_pink',
  fc > -1.5 & fc < 0 & padj <= 0.1 ~ 'dn_blue',
  fc > -1.5 & fc < 0 & padj >= 0.1 ~'dn_grey',
  fc > 1.5 & padj <= 0.1 ~ 'up_red',
  fc > 1.5 & padj >= 0.1 ~ 'up_pink',
  fc < 1.5 & fc > 0 & padj <= 0.1 ~ 'up_blue',
  fc < 1.5 & fc > 0 & padj >= 0.1 ~ 'up_grey',
  TRUE ~ 'others'
)) %>% janitor::tabyl(new) %>%
  janitor::adorn_totals()

     new    n percent
 dn_blue   20   0.020
 dn_grey  127   0.127
 dn_pink  328   0.328
  dn_red   44   0.044
 up_blue   18   0.018
 up_grey  127   0.127
 up_pink  296   0.296
  up_red   40   0.040
   Total 1000   1.000

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