Question

我正在尝试执行以下操作：

从{{1}的列ScanNo和Intensity中获取值，如果df1的值满足m/z到对应的df1['m/z'] >= 126.126226) & df1['m/z'] <= 126.129226和ScanNo中的TMT126列；
从df2的{{1}}和ScanNo列中获取值，如果Intensity的值满足df1，则将{{1 }和{{1}中的m/z到df1['m/z'] >= 127.123261) & df1['m/z'] <= 127.126261中相应的ScanNo和Intensity列中。
等

df1

df1

df2

ScanNo

有人知道如何使用R 来做到这一点吗？谢谢！

Answer 1

这是一项功能，可帮助您处理质谱数据。它使用dplyr函数。

library(dplyr)
select_scans <- function(data, mz_min, mz_max) {   
  data %>%  # setting up the pipe
    mutate(across(everything(), as.numeric)) %>% # convert all columns to numeric if needed
    filter(between(`m_z`, mz_min, mz_max)) %>%  # filtering only the m/z values you want
    select(ScanNo, Intensity) %>% # keeping only the columns you want
    rename(paste0("TM", round(average(mz_min, mz_max), 0)) = Intensity)
    # rename the intensity column like you want
}

所以，您运行

df126 <- select_scans(df1, 126.126226, 126.129226)

Answer 2

在内部使用within和ifelse。如果值不在范围内，则可能需要NA。我创建了一个简化的m.z列进行演示。

df2 <- within(df1, {
  TMT126 <- ifelse(m.z >= 1 & m.z <= 2, m.z, NA)
  TMT127 <- ifelse(m.z >= 3 & m.z <= 4, m.z, NA)
  TMT128 <- ifelse(m.z >= 5 & m.z <= 6, m.z, NA)
  rm(m.z, Intensity)
  })
df2
#    ScanNo TMT128 TMT127 TMT126
# 1       3     NA     NA      2
# 2       3     NA      3     NA
# 3       3     NA      3     NA
# 4       3     NA     NA      2
# 5       3     NA      4     NA
# 6       5     NA      4     NA
# 7       5      6     NA     NA
# 8       5      6     NA     NA
# 9       5      5     NA     NA
# 10      7      5     NA     NA
# 11      9     NA      4     NA
# 12     13     NA     NA      2
# 13     13     NA     NA      2
# 14     13      6     NA     NA
# 15     13     NA      4     NA
# 16     16      5     NA     NA
# 17     16     NA     NA      2
# 18     16     NA     NA      1
# 19     16     NA      4     NA
# 20     19     NA      4     NA

数据：

df1 <- structure(list(ScanNo = c(3L, 3L, 3L, 3L, 3L, 5L, 5L, 5L, 5L, 
7L, 9L, 13L, 13L, 13L, 13L, 16L, 16L, 16L, 16L, 19L), m.z = c(2L, 
3L, 3L, 2L, 4L, 4L, 6L, 6L, 5L, 5L, 4L, 2L, 2L, 6L, 4L, 5L, 2L, 
1L, 4L, 4L), Intensity = c(499.050819190312, 502.115755613237, 
498.921830630967, 500.373553890647, 498.659124958938, 500.670703826751, 
499.295448634045, 499.948336887528, 499.49054987242, 500.160221846888, 
500.036135738485, 500.946913174943, 500.580928969496, 498.996895445679, 
496.507093594431, 500.788140622824, 500.167440904356, 499.120163471469, 
497.046420199033, 499.682652479155)), row.names = c(NA, -20L), class = "data.frame")

如果满足某些条件，则将数据框1中的值选择为新数据框

2 个答案: