按不同列比较数据

按不同列比较数据

我有一个文件(制表符分隔)包含具有不同列数的行。像这样:

Bin_37:_Pelotomaculum_sp._DTU098    GH3 GH57    GH15    GH18    GT2 GT4 GT28                                                                        
Bin_45_1:_Thiopseudomonas_denitrificans GH3 GH57    GT2 GT9 CBM48                                                                               
...

My question is: how can I generate another file (tsv) containing the comparision of rows by column where the data are organized. Missing values are filled up with NA. For example, like this:

Bin_37:_Pelotomaculum_sp._DTU098 GH3 GH57 GH15 GH18 GT2 GT4 GT28 NA NA
Bin_45_1:_Thiopseudomonas_denitrificans GH3 GH57 NA NA GT2 NA NA GT9 CBM48
...

答案1

也许对于超大文件来说不是最有效的,但是工作版本。

输入文件文件1:

Bin_37:_Pelotomaculum_sp._DTU098        GH3     GH57    GH15    GH18    GT2     GT4     GT28
Bin_45_1:_Thiopseudomonas_denitrificans GH3     GH57    GT2     GT9     CBM48
Bin_99:_to_make_sure_no_columns_is_ok

脚本(/bin/sh 或 /bin/bash):

#!/bin/sh
F="file1";
COLS=$(cat "${F}"|sed 's/^[^\t]*//g;s/\t/\n/g'|sort|uniq|xargs);
# list of all available unique columns in SORTED order
echo "All avaiulable columns: [${COLS}]";
echo
# reading from the file line by line
cat "${F}"|while read L; do
  # assign to A the first column
  A=$(echo "${L}"|cut -d' ' -f1);
  # if A is not empty
  [ -n "${A}" ] && 
  {
    # take one by one all possible column values
    for C in ${COLS}; do
      # if the taken line has such column, add it to A,
      # otherwise add to A NA
      echo "${L} "|grep "\s${C}\s" >/dev/null && 
        A="$A"$'\t'"${C}" || 
        A="$A"$'\tNA'; 
    done;
    # print result line
    echo "${A}";
  };
done

输出

All avaiulable columns: [CBM48 GH15 GH18 GH3 GH57 GT2 GT28 GT4 GT9]

Bin_37:_Pelotomaculum_sp._DTU098        NA      GH15    GH18    GH3     GH57    GT2     GT28    GT4     NA
Bin_45_1:_Thiopseudomonas_denitrificans CBM48   NA      NA      GH3     GH57    GT2     NA      NA      GT9
Bin_99:_to_make_sure_no_columns_is_ok   NA      NA      NA      NA      NA      NA      NA      NA      NA

相同(开头没有可用列的列表)作为一个班轮

F="file1"; COLS=$(cat "${F}"|sed 's/^[^\t]*//g;s/\t/\n/g'|sort|uniq|xargs); cat "${F}"|while read L; do A=$(echo "${L}"|cut -d' ' -f1); [ -n "${A}" ] && { for C in ${COLS}; do echo "${L} "|grep "\s${C}\s" >/dev/null && A="$A"$'\t'"${C}" || A="$A"$'\tNA'; done; echo "${A}"; }; done

更新。优化更高效的版本,基于评论中的建议(需要/bin/bash):

F="file1"; IFS=$'\n'; COLS=($(sed 's/^[^\t]*//g;s/\t/\n/g' "${F}"|sort -u)); while read -r L; do A="${L%%$'\t'*}"; [ -n "${A}" ] && for C in ${COLS[@]}; do [[ "${L}"$'\t' == *$'\t'"${C}"$'\t'* ]] && A="$A"$'\t'"${C}" || A="$A"$'\tNA'; done && echo "${A}"; done <${F}; IFS=' '

答案2

与到目前为止的所有其他答案一样,这不会产生您从提供的输入中提供的预期输出,但如果您的输入实际上包含空的制表符分隔字段,那么这将用NAs 填充这些字段:

$ cat tst.awk
BEGIN { FS=OFS="\t" }
NR == FNR {
    gsub(/\t+$/,"")
    maxNF = (NF>maxNF ? NF : maxNF)
    next
}
{
    for (i=1; i<=maxNF; i++) {
        printf "%s%s", ($i == "" ? "NA" : $i), (i < maxNF ? OFS : ORS)
    }
}

$ awk -f tst.awk file file
Bin_37:_Pelotomaculum_sp._DTU098        GH3     GH57    GH15    GH18    GT2     GT4     GT28
Bin_45_1:_Thiopseudomonas_denitrificans GH3     GH57    GT2     GT9     CBM48   NA      NA

答案3

如果您的输入文件是制表符分隔的,您可以使用此 GNU awk 脚本:

awk 'BEGIN{RS="[\t\n]"} !NF{$1="NA"} {printf "%s%s", $0, RT}' file

记录分隔符RS设置为制表符或换行符,以便获取 中的字段数NF

如果NF为空,意味着两个选项卡之间没有单词,则NA添加该字符串。

该脚本使用记录终止符RT(a\t或 a \n)打印结果记录。

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