计算 DF (R) 中每 N 行的 Z 分数

Question

Hi I have a df that has variables as the columns and rows are time. The element of each intersection is a count.

      Var_1 Var_2 Var_3 
Time_1 5     4      5
Time_2 4     19     4 
Time_3 2     2     87

This df has a lot of rows (> 30,000)

How can I calculate Z scores for every 20 rows? Thanks in advance! <3

Answer 1

Here is an answer that uses dplyr::summarise() to calculate means and standard deviations, then we merge them with the original data and use mutate() to calculate the z-scores. We'll illustrate the single variable case, but it can be extended to handle multiple variables.

Given the ambiguity of the original question, we assume the Time- column is structured in groups of 20, which allows us to use it as the main grouping variable for the solution. That is, there are 20 observations at Time-1 , another 20 at Time-2 , etc.

If the requirement is to create groups of 20 rows based on consecutive row identifiers, the solution can easily be modified to add a grouping variable to represent sets of 20 rows.

# simulate some data
y <- rpois(20000,3) # simulate counts 
TimeVal <- paste0(rep("Time-",20000),
                  rep(1:1000,20))

data <-data.frame(TimeVal,y,stringsAsFactors = FALSE)
library(dplyr)
result <- data %>% group_by(TimeVal) %>% summarise(ybar = mean(y),
                                                        stDev = sd(y)) %>%
               full_join(data,.,) %>% mutate(.,zScore = (y - ybar) / stDev)
head(result)

...and the output:

> head(result)
  TimeVal y ybar    stDev      zScore
1  Time-1 6 3.45 1.276302  1.99795938
2  Time-2 2 2.95 1.700619 -0.55862010
3  Time-3 2 3.20 1.908430 -0.62878909
4  Time-4 3 3.10 1.916686 -0.05217339
5  Time-5 2 3.10 1.447321 -0.76002513
6  Time-6 2 3.30 1.809333 -0.71849700
> 

Extending the solution: z-scores for multiple columns

To solve for multiple columns in the original input data frame, first we create a long form tidy data frame with tidyr::pivot_longer) , calculate means and standard deviations, merge them with the narrow data and calculate z-scores.

Converting the input data to a long form tidy data frame allows us to use the original column names in a dplyr::by_group() , eliminating a lot of code that would be otherwise required to calculate the z-scores for each column in the original data.

library(tidyr)
set.seed(95014) # set seed to make results reproducible 
y2 <- rpois(20000,8)
y3 <- rpois(20000,15)    
data <- data.frame(TimeVal,y,y2,y3,stringsAsFactors = FALSE)

# convert to narrow format tidy, calculate means, sds, and zScores
longData <- data %>% 
  group_by(TimeVal) %>%
  pivot_longer(-TimeVal, 
      names_to = "variable",                                                         
      values_to = "value")  
result <- longData %>% 
  group_by(TimeVal,variable) %>% 
  summarise(avg = mean(value), stDev = sd(value)) %>%
  full_join(longData,.) %>% 
  mutate(.,zScore = (value - avg) / stDev)
head(result)

...and the output:

> head(result)
# A tibble: 6 x 6
# Groups:   TimeVal [2]
  TimeVal variable value   avg stDev zScore
  <chr>   <chr>    <int> <dbl> <dbl>  <dbl>
1 Time-1  y            6  3.45  1.28  2.00 
2 Time-1  y2          13  8.7   2.23  1.93 
3 Time-1  y3          20 16.4   5.25  0.686
4 Time-2  y            2  2.95  1.70 -0.559
5 Time-2  y2           6  8.2   2.89 -0.760
6 Time-2  y3          12 14.8   3.34 -0.852
>