dplyr函数可合并重复的数据，删除丢失的数据并维护冲突的数据 [英] dplyr function to combine repeated data, remove missing data, and maintain conflicting data

问题描述

我有一个数据集，其中某些条目不止一次出现，有时其中一个条目中的数据丢失，而另一个条目中的数据丢失，有时完全丢失的数据，有时数据冲突.

I have a dataset where some entries appear more than once, sometimes with data in one entry that is missing in another, sometimes with completely missing data, and sometimes with conflicting data:

# A tibble: 9 x 4 
  ID    name    age   fsm
  <chr> <chr> <dbl> <dbl>
1 0001  Peter    13    NA
2 0001  NA       13     1
3 0001  Barry    13     1
4 0002  Jane     13     1
5 0002  Jane     NA     1
6 0003  Billy    12     0
7 0003  Billy    12     1
8 0004  Sally    12    NA
9 0004  Sally    12    NA

我想将条目与丢失的数据(例如0001和0002)组合起来，维护有冲突的数据(0003)，并维护将NA翻倍的数据(0004).要产生这个:

I would like to combine entries with missing data (e.g. 0001 and 0002), maintain conflicting data (0003) and maintain data that double NA (0004). To produce this:

# A tibble: 9 x 4 
  ID    name    age   fsm
  <chr> <chr> <dbl> <dbl>
1 0001  Peter    13     1
2 0001  Barry    13     1
3 0002  Jane     13     1
4 0003  Billy    12     0
5 0003  Billy    12     1
6 0004  Sally    12    NA

在此基础上:如何将重复的行与缺少的字段R合并

我写了以下内容:

tmp %>%
group_by(ID) %>%
summarise(across(everything(), ~ ifelse(length(na.omit(.x)) == 0, NA, na.omit(.x)))) %>%
distinct()

但它输掉了两次输入0003

But it loses the double entry, 0003

   ID    name    age   fsm
   <chr> <chr> <dbl> <dbl>
 1 0001  Peter    13     1
 2 0002  Jane     13     1
 3 0003  Billy    12     0
 4 0004  Sally    12    NA

数据作为吞吐量:

structure(list(ID = c("0001", "0001", "0001", "0002", "0002", 
"0003", "0003", "0004", "0004"), name = c("Peter", NA, "Barry", 
"Jane", "Jane", "Billy", "Billy", "Sally", "Sally"), age = c(13, 
13, 13, 13, NA, 12, 12, 12, 12), fsm = c(NA, 1, 1, 1, 1, 0, 1, 
NA, NA)), class = c("spec_tbl_df", "tbl_df", "tbl", "data.frame"
), row.names = c(NA, -9L), spec = structure(list(cols = list(
ID = structure(list(), class = c("collector_character", "collector"
)), name = structure(list(), class = c("collector_character", 
"collector")), age = structure(list(), class = c("collector_double", 
"collector")), fsm = structure(list(), class = c("collector_double", 
"collector"))), default = structure(list(), class = c("collector_guess", 
"collector")), skip = 1L), class = "col_spec"))

推荐答案

您是否正在寻找这种解决方案?

Are you looking for this kind of solution?

library(tidyverse)

df %>% 
  group_by(ID, name, age, fsm) %>% 
  mutate(dupe = n()>1) %>% 
  group_by(ID) %>%
  dplyr::summarise(across(everything(), ~ ifelse(length(na.omit(.x)) == 0  & dupe == TRUE, NA, na.omit(.x)))) %>%
  distinct() %>% 
  select(-dupe)

输出:

  ID    name    age   fsm
  <chr> <chr> <dbl> <dbl>
1 0001  Peter    13     1
2 0001  Barry    13     1
3 0002  Jane     13     1
4 0003  Billy    12     0
5 0003  Billy    12     1
6 0004  Sally    12    NA

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