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，这很容易：

$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

在朱莉娅中，

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

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

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

为此，我需要 2 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" 

是否有一步解决方案或更好（更快）的方法？

只是为了好玩，所有这些有趣的答案的基准！ （使用 Julia 5.0 更新）

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

我将基准更改为执行 5 次 @time，并在此处发布执行 5 次 @time 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

就像@daycaster 建议的那样，您可以使用s"\1" 作为替换字符串来支持捕获组。问题是特殊的s"" 字符串语法不支持像\n 这样的特殊字符。您可以通过手动构造一个SubstitutionString 对象来解决此问题，但是您需要转义\1 中的\：

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

Answer 2

如果速度是个问题，最好避免使用较重的工具，例如正则表达式，并尝试只在底层完成工作，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 μs (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 μs (0.00% GC)
  median time:      2.29 μs (0.00% GC)
  mean time:        2.43 μs (3.85% GC)
  maximum time:     531.31 μs (98.95% GC)

只有 2.5 倍的加速。 replace函数实现的非常好！

另一种不用正则表达式的方法是

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

这不是那么快（慢 10 倍，在我的机器上没有内存分配损失），但非常易读。

Answer 3

类似：

julia> split(replace(seq, r"(.{10})", s"\1 "))
6-element Array{SubString{String},1}:
 "MSKNKSPLLN"
 "ESEKMMSEML"
 "PMKVSQSKLN"
 "YEEKVYIPTT"
 "IRNRKQHCFR"
 "RFFPYIALFQ"

如果你想要它作为一个字符串，使用join():

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

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

Answer 4

我不知道你如何使用 REGEX，但我认为它可以解决你的问题：

a = "oiaoueaoeuaoeuaoeuaoeuaoteuhasonetuhaonetuahounsaothunsaotuaosu"
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)