使用Julia在字符串中每10个字符插入一个换行符

Question

我想在蛋白质序列中每10个字符插入一个换行符:

seq="MSKNKSPLLNESEKMMSEMLPMKVSQSKLNYEEKVYIPTTIRNRKQHCFRRFFPYIALFQ"

在Perl中,它非常简单:

$seq=~s/(.{10})/$1\n/g ; # does the job!

perl -e '$seq="MSKNKSPLLNESEKMMSEMLPMKVSQSKLNYEEKVYIPTTIRNRKQHCFRRFFPYIALFQ"; $seq=~s/(.{10})/$1\n/g; print $seq'
MSKNKSPLLN
ESEKMMSEML
PMKVSQSKLN
YEEKVYIPTT
IRNRKQHCFR
RFFPYIALFQ

在朱莉娅,

replace(seq, r"(.{10})" , "\n")

不起作用,因为我不知道如何获取捕获组(.{10})并将其替换为自己+"\n"

julia> replace(seq, r"(.{10})" , "\n")
"\n\n\n\n\n\n"

所以要做到这一点,我需要两个步骤:

    julia> a=matchall(r"(.{1,10})" ,seq)
    6-element Array{SubString{UTF8String},1}:
     "MSKNKSPLLN"
     "ESEKMMSEML"
     "PMKVSQSKLN"
     "YEEKVYIPTT"
     "IRNRKQHCFR"
     "RFFPYIALFQ"

    julia> b=join(a, "\n")
    "MSKNKSPLLN\nESEKMMSEML\nPMKVSQSKLN\nYEEKVYIPTT\nIRNRKQHCFR\nRFFPYIALFQ"

    julia> println(b)
    MSKNKSPLLN
    ESEKMMSEML
    PMKVSQSKLN
    YEEKVYIPTT
    IRNRKQHCFR
    RFFPYIALFQ

# Caution :    
a=matchall(r"(.{10})" ,seq) # wrong if seq is not exactly a multiple of 10 !

julia> seq
"MSKNKSPLLNESEKMMSEMLPMKVSQSKLNYEEKVYIPTTIRNRKQHCFRRFFPYIAL"

julia> matchall(r"(.{10})" ,seq)
5-element Array{SubString{UTF8String},1}:
"MSKNKSPLLN"
"ESEKMMSEML"
"PMKVSQSKLN"
"YEEKVYIPTT"
"IRNRKQHCFR"

julia> matchall(r"(.{1,10})" ,seq)
6-element Array{SubString{UTF8String},1}:
"MSKNKSPLLN"
"ESEKMMSEML"
"PMKVSQSKLN"
"YEEKVYIPTT"
"IRNRKQHCFR"
"RFFPYIAL" 

有一步解决方案还是更好(更快)的方式？

只是为了有趣的基准与所有这些有趣的答案!(更新了julia 5.0)

function loop(a)
 last = 0
 #create the interval, in your case 10
 salt = 10
 #iterate in string (starts in the 10th value, don't forget julia use 1 to first index)
 for i in salt:salt+1:length(a)
    # replace the string for a new one with '\n'
    a = string(a[1:i], '\n', a[i+1:length(a)])
    last = Int64(i)
 end
 # replace the rest
 a = string(a[1:length(a) - last % salt + 1], '\n', a[length(a) - last % salt + 2:length(a)])
 println(a)
end

function regex1(seq)
  a=matchall(r"(.{1,10})" ,seq)
  b=join(a, "\n")
  println(b)
end

function regex2(seq)
  a=join(split(replace(seq, r"(.{10})", s"\1 ")), "\n")
  println(a)
end

function regex3(seq)
  a=replace(seq, r"(.{10})", Base.SubstitutionString("\\1\n"))
  a= chomp(a) # because there is a new line at the end
  println(a)
end

function intrapad(seq::String)
  buf = IOBuffer((length(seq)*11)>>3) # big enough buffer
  for i=1:10:length(seq)
    write(buf,SubString(seq,i,i+9),'\n')
  end
  #return
  print(takebuf_string(buf))
end

function join_substring(seq)
  a=join((SubString(seq,i,i+9) for i=1:10:length(seq)),'\n')
  println(a)
end

seq="MSKNKSPLLNESEKMMSEMLPMKVSQSKLNYEEKVYIPTTIRNRKQHCFRRFFPYIALFQ"

for i = 1:5
  println("loop :")
  @time loop(seq)
  println("regex1 :")
  @time regex1(seq)
  println("regex2 :")
  @time regex2(seq)
  println("regex3 :")
  @time regex3(seq)
  println("intrapad :")
  @time intrapad(seq)
  println("join substring :")
  @time join_substring(seq)
end

我更改了基准测试以执行5次@time并在此处发布@time执行5次后的结果:

loop :
MSKNKSPLLN
ESEKMMSEML
PMKVSQSKLN
YEEKVYIPTT
IRNRKQHCFR
RFFPYIA
LFQ
  0.000013 seconds (53 allocations: 3.359 KB)
regex1 :
MSKNKSPLLN
ESEKMMSEML
PMKVSQSKLN
YEEKVYIPTT
IRNRKQHCFR
RFFPYIALFQ
  0.000013 seconds (49 allocations: 1.344 KB)
regex2 :
MSKNKSPLLN
ESEKMMSEML
PMKVSQSKLN
YEEKVYIPTT
IRNRKQHCFR
RFFPYIALFQ
  0.000017 seconds (47 allocations: 1.703 KB)
regex3 :
MSKNKSPLLN
ESEKMMSEML
PMKVSQSKLN
YEEKVYIPTT
IRNRKQHCFR
RFFPYIALFQ
  0.000013 seconds (31 allocations: 976 bytes)
intrapad :
MSKNKSPLLN
ESEKMMSEML
PMKVSQSKLN
YEEKVYIPTT
IRNRKQHCFR
RFFPYIALFQ
  0.000007 seconds (9 allocations: 608 bytes)
join substring :
MSKNKSPLLN
ESEKMMSEML
PMKVSQSKLN
YEEKVYIPTT
IRNRKQHCFR
RFFPYIALFQ
  0.000012 seconds (21 allocations: 800 bytes)

Intrapad现在是第一个;)

Answer 1

像@daycaster建议的那样,您可以使用s"\1"替换字符串来支持捕获组.麻烦的是特殊s""字符串语法不支持特殊字符\n.你可以通过手动构建一个SubstitutionString对象来解决这个问题,但是你需要转义\in \1:

julia> replace(seq, r"(.{10})", Base.SubstitutionString("\\1\n"))
"MSKNKSPLLN\nESEKMMSEML\nPMKVSQSKLN\nYEEKVYIPTT\nIRNRKQHCFR\nRFFPYIALFQ\n"

Answer 2

如果速度是一个问题,最好避免使用较重的工具,例如正则表达式,并尽量让低级别的工作完成,如下所示:

function intrapad(seq::String)
  buf = IOBuffer((length(seq)*11)>>3) # big enough buffer
  for i=1:10:length(seq)
    write(buf,SubString(seq,i,i+9),'\n')
  end
  return takebuf_string(buf)
end

速度来自使用IOBuffer和SubStrings最小化分配.使用BenchmarkTools包我们有:

julia> @benchmark intrapad(seq)
BenchmarkTools.Trial: 
  memory estimate:  624.00 bytes
  allocs estimate:  10
  minimum time:     729.00 ns (0.00% GC)
  median time:      767.00 ns (0.00% GC)
  mean time:        862.99 ns (7.84% GC)
  maximum time:     26.86 ?s (96.21% GC)

julia> @benchmark replace(seq, r"(.{10})", Base.SubstitutionString("\\1\n"))
BenchmarkTools.Trial: 
  memory estimate:  720.00 bytes
  allocs estimate:  26
  minimum time:     2.18 ?s (0.00% GC)
  median time:      2.29 ?s (0.00% GC)
  mean time:        2.43 ?s (3.85% GC)
  maximum time:     531.31 ?s (98.95% GC)

只有2.5倍的加速.该replace功能是很好的实现!

没有正则表达式的另一种方法是

join((SubString(seq,i,i+9) for i=1:10:length(seq)),'\n')

这不是那么快(10倍慢,我的机器没有内存分配惩罚),但非常可读.