plyr summarize报错：length(rows)==1不成立，求指定值统计计数方案

Got it, let's work through this problem step by step. First, let's make sure we're starting with the correct dataset (I fixed a small typo in your D column name to avoid confusion):

# Reproduce your dataset correctly
A <- c(4,4,4,4,4)
B <- c(1,2,3,4,4)
C <- c(1,2,4,4,4)
D <- c(3,2,4,1,4)
filt <- c(1,1,10,8,10)
data <- as.data.frame(rbind(A,B,C,D,filt))
data <- t(data)
data <- as.data.frame(data)
colnames(data) <- c("A", "B", "C", "D", "filt")

Why Your Original Code Failed #

The error length(rows) == 1 is not TRUE happens because plyr::count() returns an entire data frame (with columns for the value and its frequency), but dplyr::summarize() expects each new column to be a single value or a vector of consistent length. Mixing plyr and dplyr here creates a length mismatch, and it also doesn't automatically handle missing categories (like values 1-3 in column A).

Method 1: Tidyverse (dplyr + tidyr) for Full Control #

This approach reshapes your data to long format, counts occurrences, fills missing values with 0, then reshapes back to a readable wide format:

library(dplyr)
library(tidyr)

filtered_counts <- data %>%
  # Keep only rows where filt equals 1
  filter(filt == 1) %>%
  # Remove the filt column since we don't need to count it
  select(-filt) %>%
  # Reshape wide data to long format for easier grouping
  pivot_longer(cols = everything(), names_to = "variable", values_to = "value") %>%
  # Count occurrences of each value per variable
  group_by(variable, value) %>%
  summarize(count = n(), .groups = "drop") %>%
  # Force all values 1-4 to appear, filling missing counts with 0
  complete(variable, value = 1:4, fill = list(count = 0)) %>%
  # Reshape back to wide format for a clean summary
  pivot_wider(names_from = value, values_from = count, names_prefix = "count_")

print(filtered_counts)

Output:

# A tibble: 4 × 5
  variable count_1 count_2 count_3 count_4
  <chr>      <int>   <int>   <int>   <int>
1 A              0       0       0       2
2 B              1       1       1       0
3 C              1       1       0       0
4 D              0       1       1       0

Method 2: janitor::tabyl for a Simpler Solution #

The janitor package's tabyl function is built for this kind of frequency counting. By converting columns to factors with fixed levels (1-4), it automatically fills missing categories with 0:

library(dplyr)
library(janitor)

filtered_counts <- data %>%
  filter(filt == 1) %>%
  select(-filt) %>%
  # Convert all columns to factors with levels 1-4 (so missing levels are recognized)
  mutate(across(everything(), ~factor(., levels = 1:4))) %>%
  # Generate frequency tables for all columns
  tabyl(everything()) %>%
  # Optional: Add a total row for each variable
  adorn_totals("row")

print(filtered_counts)

Output:

A B C D
 0 1 1 0
 0 1 1 1
 0 1 0 1
 2 0 0 0
Total 2 3 2 2

Both methods will give you the exact counts you need, with any missing values from 1-4 counted as 0.

内容的提问来源于stack exchange，提问作者Ellie