将ASCII字符串转换为相应小数的字符串

时间:2013-07-22 10:36:01

标签: string bash awk

我可以向您介绍一个破坏我周末的问题。我有4列生物学数据

@ID:::12345/1 ACGACTACGA text !"#$%vwxyz  
@ID:::12345/2 TATGACGACTA text :;<=>?VWXYZ

我想使用awk编辑第一列来替换字符:和/ with -
我想用逗号分隔的小数字串转换最后一列中的字符串,这些小数字对应于每个单独的ASCII字符(任何字符,范围从ASCII 33 - 126)。

@ID---12345-1 ACGACTACGA text 33,34,35,36,37,118,119,120,121,122  
@ID---12345-2 TATGACGACTA text 58,59,60,61,62,63,86,87,88,89,90

第一部分很简单,但我坚持第二部分。我尝试过使用awk序数函数和sprintf;我只能让前者处理字符串中的第一个字符,我只能得到后者将十六进制转换为十进制而不是空格。还尝试了bash功能

$ od -t d1 test3 | awk 'BEGIN{OFS=","}{i = $1; $1 = ""; print $0}' 

但不知道如何在awk中调用此函数。 我更喜欢使用awk,因为我有一些下游操作也可以在awk中完成。

非常感谢提前

2 个答案:

答案 0 :(得分:1)

使用awk manual中的序数函数,您可以这样做:

awk -f ord.awk  --source '{
    # replace : with - in the first field
    gsub(/:/,"-",$1)

    # calculate the ordinal by looping over the characters in the fourth field
    res=ord($4)
    for(i=2;i<=length($4);i++) {
        res=res","ord(substr($4,i))
    }
    $4=res
}1' file

输出:

@ID---12345/1 ACGACTACGA text 33,34,35,36,37,118,119,120,121,122
@ID---12345/2 TATGACGACTA text 58,59,60,61,62,63,86,87,88,89,90

以下是ord.awk(取自:http://www.gnu.org/software/gawk/manual/html_node/Ordinal-Functions.html

# ord.awk --- do ord and chr

# Global identifiers:
#    _ord_:        numerical values indexed by characters
#    _ord_init:    function to initialize _ord_



BEGIN    { _ord_init() }

function _ord_init(    low, high, i, t)
{
    low = sprintf("%c", 7) # BEL is ascii 7
    if (low == "\a") {    # regular ascii
        low = 0
        high = 127
    } else if (sprintf("%c", 128 + 7) == "\a") {
        # ascii, mark parity
        low = 128
        high = 255
    } else {        # ebcdic(!)
        low = 0
        high = 255
    }

    for (i = low; i <= high; i++) {
        t = sprintf("%c", i)
        _ord_[t] = i
    }
}

function ord(str, c)
{
    # only first character is of interest
    c = substr(str, 1, 1)
    return _ord_[c]
}

function chr(c)
{
    # force c to be numeric by adding 0
    return sprintf("%c", c + 0)
}

如果您不想包含整个ord.awk,可以这样做:

awk 'BEGIN{ _ord_init()}
function _ord_init(low, high, i, t)
{
    low = sprintf("%c", 7) # BEL is ascii 7
    if (low == "\a") {    # regular ascii
        low = 0
        high = 127
    } else if (sprintf("%c", 128 + 7) == "\a") {
        # ascii, mark parity
        low = 128
        high = 255
    } else {        # ebcdic(!)
        low = 0
        high = 255
    }

    for (i = low; i <= high; i++) {
        t = sprintf("%c", i)
        _ord_[t] = i
    }
}
{
    # replace : with - in the first field
    gsub(/:/,"-",$1)

    # calculate the ordinal by looping over the characters in the fourth field
    res=_ord_[substr($4,1,1)]
    for(i=2;i<=length($4);i++) {
        res=res","_ord_[substr($4,i,1)]
    }
    $4=res
}1' file

答案 1 :(得分:0)

Perl soltuion:

perl -lnae '$F[0] =~ s%[:/]%-%g; $F[-1] =~ s/(.)/ord($1) . ","/ge; chop $F[-1]; print "@F";' < input

第一个替换用短划线替换第一个字段中的:/,第二个替换用其ord和逗号替换最后一个字段中的每个字符,chop删除最后一个逗号。

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