Question

我有一个包含许多列的数据。例如，这是三列

df<-structure(list(V1 = structure(c(5L, 1L, 7L, 3L, 2L, 4L, 6L, 6L
), .Label = c("CPSIAAAIAAVNALHGR", "DLNYCFSGMSDHR", "FPEHELIVDPQR", 
"IADPDAVKPDDWDEDAPSK", "LWADHGVQACFGR", "WGEAGAEYVVESTGVFTTMEK", 
"YYVTIIDAPGHR"), class = "factor"), V2 = structure(c(5L, 2L, 
7L, 3L, 4L, 6L, 1L, 1L), .Label = c("", "CPSIAAAIAAVNALHGR", 
"GCITIIGGGDTATCCAK", "HVGPGVLSMANAGPNTNGSQFFICTIK", "LLELGPKPEVAQQTR", 
"MVCCSAWSEDHPICNLFTCGFDR", "YYVTIIDAPGHR"), class = "factor"), 
    V3 = structure(c(4L, 3L, 2L, 4L, 3L, 1L, 1L, 1L), .Label = c("", 
    "AVCMLSNTTAIAEAWAR", "DLNYCFSGMSDHR", "FPEHELIVDPQR"), class = "factor")), .Names = c("V1", 
"V2", "V3"), class = "data.frame", row.names = c(NA, -8L))

- 第一栏，我们不看任何其他专栏，我们只计算有多少个字符串并保留唯一的字符串

第二栏，我们保持唯一，我们也删除第一栏中已有的那些
第三列，我们保持唯一，我们删除第一列和第二列中的字符串

这与我们一样多的列继续

例如，对于这些数据，我们将有以下

 Column 1              Column 2                    Column 3
LWADHGVQACFGR
CPSIAAAIAAVNALHGR     LLELGPKPEVAQQTR              AVCMLSNTTAIAEAWAR
YYVTIIDAPGHR          GCITIIGGGDTATCCAK 
FPEHELIVDPQR          HVGPGVLSMANAGPNTNGSQFFICTIK   
DLNYCFSGMSDHR         MVCCSAWSEDHPICNLFTCGFDR   
IADPDAVKPDDWDEDAPSK     
WGEAGAEYVVESTGVFTTMEK

Answer 1

以下是tidyverse，

的解决方案

library(tidyverse)

df1 <- df %>% 
 gather(var, string) %>% 
 filter(string != '' & !duplicated(string)) %>% 
 group_by(var) %>% 
 mutate(cnt = seq(n())) %>% 
 spread(var, string) %>%
 select(-cnt)

哪个给出了

# A tibble: 7 x 4
    cnt                    V1                          V2                V3
* <int>                 <chr>                       <chr>             <chr>
1     1         LWADHGVQACFGR             LLELGPKPEVAQQTR AVCMLSNTTAIAEAWAR
2     2     CPSIAAAIAAVNALHGR           GCITIIGGGDTATCCAK              <NA>
3     3          YYVTIIDAPGHR HVGPGVLSMANAGPNTNGSQFFICTIK              <NA>
4     4          FPEHELIVDPQR     MVCCSAWSEDHPICNLFTCGFDR              <NA>
5     5         DLNYCFSGMSDHR                        <NA>              <NA>
6     6   IADPDAVKPDDWDEDAPSK                        <NA>              <NA>
7     7 WGEAGAEYVVESTGVFTTMEK                        <NA>              <NA>

您可以使用colSums来获取字符串数量

colSums(!is.na(df1))
#V1 V2 V3 
# 7  4  1

通过基本R的类似方法将保存列表中的字符串，

df[] <- lapply(df, as.character)
d1 <- stack(df)
d1 <- d1[d1$values != '' & !duplicated(d1$values),]
l1 <- unstack(d1, values ~ ind)

lengths(l1)
#V1 V2 V3 
# 7  4  1

Answer 2

基础R解决方案。 df2是最终输出。

# Convert to character
L1 <- lapply(df, as.character)
# Get unique string
L2 <- lapply(L1, unique)
# Remove ""
L3 <- lapply(L2, function(vec){vec <- vec[!(vec %in% "")]})

# Use for loop to remove non-unique string from previous columns
for (i in 2:length(L3)){
  previous_vec <- unlist(L3[1:(i - 1)])
  current_vec <- L3[[i]]
  L3[[i]] <- current_vec[!(current_vec %in% previous_vec)]
}

# Get the maximum column length
max_num <- max(sapply(L3, length))

# Append "" to each column
L4 <- lapply(L3, function(vec){vec <- c(vec, rep("", max_num - length(vec)))})

# Convert L4 to a data frame
df2 <- as.data.frame(do.call(cbind, L4))

如何计算和删除列之间的类似字符串

2 个答案: