在 R 中消除多个连续的字符串下单元实例答案

【问题标题】：Eliminating more than one succeeding instance of a string down cells in R在 R 中消除多个连续的字符串下单元实例
【发布时间】：2020-08-06 00:27:08
【问题描述】：

我对 R 比较陌生。我有一个包含 500 万个观察值和 1 个变量的数据框，看起来像这样：

PMID- 28524368 PMID- 28504342 PMID- 28501042 RN - 4964P6T9RB (Aldosterone) RN - EC 3.4.23.15 (Renin) RN - RWP5GA015D (Potassium) MH - Adrenal Cortex Neoplasms/*diagnostic imaging/pathology/surgery MH - Adrenocortical Adenoma/*diagnostic imaging/pathology/surgery MH - Aldosterone/blood MH - Humans PMID- 28523858 PMID- 28517030 PMID- 28513869 MH - Hyperaldosteronism/*complications MH - Hypertension/*etiology MH - Male MH - Middle Aged MH - Potassium/blood PMID- 28494487 PMID- 28493475 MH - Renin/blood MH - Tomography, X-Ray Computed

但是，我只希望连续有 1 个 PMID，第一个 - 其余的 PMID 应该被删除，导致数据帧看起来像：

PMID- 28524368 RN - 4964P6T9RB (Aldosterone) RN - EC 3.4.23.15 (Renin) RN - RWP5GA015D (Potassium) MH - Adrenal Cortex Neoplasms/*diagnostic imaging/pathology/surgery MH - Adrenocortical Adenoma/*diagnostic imaging/pathology/surgery MH - Aldosterone/blood MH - Humans PMID- 28523858 MH - Hyperaldosteronism/*complications MH - Hypertension/*etiology MH - Male MH - Middle Aged MH - Potassium/blood PMID- 28494487 MH - Renin/blood MH - Tomography, X-Ray Computed

请指教。我尝试使用：

# remove excessive PMIDs
for (i in nrow(original_reduced))
{
  if (substr(original_reduced[i, 1], 1, 4) == "PMID")
  {
    if (substr(original_reduced[i+1, 1], 1, 4) == "PMID" && i != nrow(original_reduced)) # if next row is also PMID
    {
      original_reduced <- original_reduced[-c(i+1), ] # delete entry after
    }
  }
}

但我收到了这个错误：

Error in if (substr(original_reduced[i + 1, 1], 1, 4) == "PMID") { : missing value where TRUE/FALSE needed

即使我的数据框中没有 NA。

谢谢。

【问题讨论】：

标签： r string dataframe substr

【解决方案1】：

这是一个可行的解决方案。代码解释见 cmets

 df<-structure(list(V1 = c("PMID- 28524368", "PMID- 28504342", "PMID- 28501042", 
"RN - 4964P6T9RB (Aldosterone)", "RN - EC 3.4.23.15 (Renin)", 
"RN - RWP5GA015D (Potassium)", "MH - Adrenal Cortex Neoplasms/*diagnostic imaging/pathology/surgery", 
"MH - Adrenocortical Adenoma/*diagnostic imaging/pathology/surgery", 
"MH - Aldosterone/blood", "MH - Humans", "PMID- 28523858", "PMID- 28517030", 
"PMID- 28513869", "MH - Hyperaldosteronism/*complications", "MH - Hypertension/*etiology", 
"MH - Male", "MH - Middle Aged", "MH - Potassium/blood", "PMID- 28494487", 
"PMID- 28493475", "MH - Renin/blood", "MH - Tomography, X-Ray Computed"
)), .Names = "V1", row.names = c(NA, -22L), class = "data.frame")

library(dplyr)

#Add flag for PMID rows
   df$pmid<-grepl("^PMID", df$V1)
#find rows of where n == n+1
   matches<-df$pmid==lag(df$pmid)
#find rows equal to previous row and is a PMID row
   toremove<-which(matches==TRUE & df$pmid==TRUE)
#remove rows
   df<-df[-toremove,]
   df$pmid<-NULL  #remove added column

【讨论】：

您的代码删除了所有的 PMID...我假设 lag 函数没有正常工作...看起来很有希望，但是
是否加载了 dplyr 库？如果使用基础包的 lag 函数，这将不起作用。
我通过将 n = 1 添加到您的 lag 函数来修复它...谢谢！ matches <-df$pmid == lag(df$pmid, n = 1)。也非常感谢您的 cmets - 作为初学者，他们真的帮助了我！

【解决方案2】：

试试这个：

df%>%mutate(number=sequence(rle(name)[['lengths']]))%>%filter((number==1 & grepl('PMID',number))|!grepl('PMID',name))%>%select(name)

【讨论】：