Question

如何根据GeneID重新格式化data.frame df1。必须根据常见的GeneID对表进行分组。我也希望突破位置

df1 =

GeneID  Common Organism Name        Position    
    3   mouse   10090   Acadm   Chr5:26082574-26089291(-)   
    3   human   9606    ACADM   Chr5:15028950-15032998(-)   
    6   mouse   10090   Acat1   Chr5:25999022-26004798(-)   
    6   human   9606    ACAT1   Chr5:15471699-15477027(-)   
    7   human   9606    NLN Chr5:26257691-26264308(+)   
    8   mouse   10090   canct1  Chr5:14910122-14914899(-)   
    9   mouse   9606    Gm10220 Chr5:25936465-25943267(-)   
    9   mouse   9606    Gm10354 Chr5:25949797-25954344(-)   
    9   mouse   9606    Gm1979  Chr5:11594913-11599784(+)   
    9   human   10090   TRIL    Chr7:28953358-28958413(-)

预期结果

Gene.ID M.Gene  M.Chr M.start   M.end      H.Gene H.Chr H.start     H.end
    3   Acadm   5   26082574    26089291    ACADM   5   15028950    15032998
    6   Acat1   5   25999022    26004798    ACAT1   5   15471699    15477027
    7   NA      NA  NA          NA           NLN    5   26257691    26264308
    8   canct1  5   14910122    14914899     NA       NA    NA      NA
    9   Gm10220 5   25936465    25943267    TRIL    7   28953358    28958413
    9   Gm10354 5   25949797    25954344    TRIL    7   28953358    28958413
    9   Gm1979  5   1159491     11599784    TRIL    7   28953358    28958413
    9   Gm21149 5   11594913    11599784    TRIL    7   28953358    28958413

Answer 1

我们可以使用devel版本的＆＃39; data.table＆＃39;即。 v1.9.5。安装说明为here。

我们更改了＆＃39; data.frame＆＃39;到＆＃39; data.table＆＃39; （setDT(df1)）。使用tstrsplit，我们将＆＃39; Position＆＃39;所有非数字字符（[^0-9]+）都可以创建新列（＆＃39; Chr＆＃39;，＆＃39; start＆＃39;，＆＃39; end＆＃39;）。

library(data.table)#v1.9.5+
DT <- setDT(df1)[, c('Chr', 'start', 'end') :=tstrsplit(Position, '[^0-9]+')[-1]]

创建一个按照＆＃39; GeneID＆＃39;分组的序列列（＆＃39; ind＆＃39;）和＃＆＃39; Common＆＃39;

DT[, ind:=1:.N, .(GeneID, Common)]

devel版本中的

dcast可以使用多个value.var列并更改“长”字样。格式为＆＃39;宽＆＃39;格式。我们可以用数据集中的非NA值替换NA值。

dcast(DT, GeneID+ind~substr(Common, 1, 1), value.var=names(DT)[c(4,6:8)])[,
 lapply(.SD, function(x) x[!is.na(x)]) , GeneID, .SDcols=h_Name:m_end]
#    GeneID h_Name  m_Name h_Chr m_Chr  h_start  m_start    h_end    m_end
#1:      3  ACADM   Acadm     5     5 15028950 26082574 15032998 26089291
#2:      6  ACAT1   Acat1     5     5 15471699 25999022 15477027 26004798
#3:      7    NLN      NA     5    NA 26257691       NA 26264308       NA
#4:      8     NA  canct1    NA     5       NA 14910122       NA 14914899
#5:      9   TRIL Gm10220     7     5 28953358 25936465 28958413 25943267
#6:      9   TRIL Gm10354     7     5 28953358 25949797 28958413 25954344
#7:      9   TRIL  Gm1979     7     5 28953358 11594913 28958413 11599784

Answer 2

使用lapply

的另一个选项

# using split method from akrun's answer
library(data.table)#v1.9.5+
DT <- setDT(df1)[, c('Chr', 'start', 'end') :=tstrsplit(Position, '[^0-9]+')[-1]]

out = setDF(Reduce(function(...) merge(..., by="GeneID", all = T),
            lapply(split(DT, DT$Common), 
            function(x) subset(x, select = -c(Common, Position, Organism)))))  

colnames(out) = gsub("x", "H", colnames(out)) 
colnames(out) = gsub("y", "M", colnames(out))

#> out
#  GeneID Name.H Chr.H  start.H    end.H  Name.M Chr.M  start.M    end.M
#1      3  ACADM     5 15028950 15032998   Acadm     5 26082574 26089291
#2      6  ACAT1     5 15471699 15477027   Acat1     5 25999022 26004798
#3      7    NLN     5 26257691 26264308    <NA>  <NA>     <NA>     <NA>
#4      8   <NA>  <NA>     <NA>     <NA>  canct1     5 14910122 14914899
#5      9   TRIL     7 28953358 28958413 Gm10220     5 25936465 25943267
#6      9   TRIL     7 28953358 28958413 Gm10354     5 25949797 25954344
#7      9   TRIL     7 28953358 28958413  Gm1979     5 11594913 11599784

如何格式化data.frame？

2 个答案: