问题
I have a data like this
>sp|Q96A73|P33MX_HUMAN Putative monooxygenase p33MONOX OS=Homo sapiens OX=9606 GN=KIAA1191 PE=1 SV=1
RNDDDDTSVCLGTRQCSWFAGCTNRTWNSSAVPLIGLPNTQDYKWVDRNSGLTWSGNDTCLYSCQNQTKGLLYQLFRNLFCSYGLTEAHGKWRCADASITNDKGHDGHRTPTWWLTGSNLTLSVNNSGLFFLCGNGVYKGFPPKWSGRCGLGYLVPSLTRYLTLNASQITNLRSFIHKVTPHR
>sp|P13674|P4HA1_HUMAN Prolyl 4-hydroxylase subunit alpha-1 OS=Homo sapiens OX=9606 GN=P4HA1 PE=1 SV=2
VECCPNCRGTGMQIRIHQIGPGMVQQIQSVCMECQGHGERISPKDRCKSCNGRKIVREKKILEVHIDKGMKDGQKITFHGEGDQEPGLEPGDIIIVLDQKDHAVFTRRGEDLFMCMDIQLVEALCGFQKPISTLDNRTIVITSHPGQIVKHGDIKCVLNEGMPIYRRPYEKGRLIIEFKVNFPENGFLSPDKLSLLEKLLPERKEVEE
>sp|Q7Z4N8|P4HA3_HUMAN Prolyl 4-hydroxylase subunit alpha-3 OS=Homo sapiens OX=9606 GN=P4HA3 PE=1 SV=1
MTEQMTLRGTLKGHNGWVTQIATTPQFPDMILSASRDKTIIMWKLTRDETNYGIPQRALRGHSHFVSDVVISSDGQFALSGSWDGTLRLWDLTTGTTTRRFVGHTKDVLSVAFSSDNRQIVSGSRDKTIKLWNTLGVCKYTVQDESHSEWVSCVRFSPNSSNPIIVSCGWDKLVKVWNLANCKLK
>sp|P04637|P53_HUMAN Cellular tumor antigen p53 OS=Homo sapiens OX=9606 GN=TP53 PE=1 SV=4
IQVVSRCRLRHTEVLPAEEENDSLGADGTHGAGAMESAAGVLIKLFCVHTKALQDVQIRFQPQL
>sp|P10144|GRAB_HUMAN Granzyme B OS=Homo sapiens OX=9606 GN=GZMB PE=1 SV=2
MQPILLLLAFLLLPRADAGEIIGGHEAKPHSRPYMAYLMIWDQKSLKRCGGFLIRDDFVLTAAHCWGSSINVTLGAHNIKEQEPTQQFIPVKRPIPHPAYNPKNFSNDIMLLQLERKAKRTRAVQPLRLPSNKAQVKPGQTCSVAGWGQTAPLGKHSHTLQEVKMTVQEDRKCES
>sp|Q9UHX1|PUF60_HUMAN Poly(U)-binding-splicing factor PUF60 OS=Homo sapiens OX=9606 GN=PUF60 PE=1 SV=1
MGKDYYQTLGLARGASDEEIKRAYRRQALRYHPDKNKEPGAEEKFKEIAEAYDVLSDPRKREIFDRYGEEGLKGSGPSGGSGGGANGTSFSYTFHGDPHAMFAEFFGGRNPFDTFFGQRNGEEGMDIDDPFSGFPMGMGGFTNVNFGRSRSAQEPARKKQDPPVTHDLRVSLEEIYSGCTKKMKISHK
>sp|Q06416|P5F1B_HUMAN Putative POU domain, class 5, transcription factor 1B OS=Homo sapiens OX=9606 GN=POU5F1B PE=5 SV=2
IVVKGHSTCLSEGALSPDGTVLATASHDGYVKFWQIYIEGQDEPRCLHEWKPHDGRPLSCLLFCDNHKKQDPDVPFWRFLITGADQNRELKMWCTVSWTCLQTIRFSPDIFSSVSVPPSLKVCLDLSAEYLILSDVQRKVLYVMELLQNQEEGHACFSSISEFLLTHPVLSFGIQVVSRCRLRHTEVLPAEEENDSLGADGTHGAGAMESAAGVLIKLFCVHTKALQDVQIRFQPQLNPDVVAPLPTHTAHEDFTFGESRPELGSEGLGSAAHGSQPDLRRIVELPAPADFLSLSSETKPKLMTPDAFMTPSASLQQITASPSSSSSGSSSSSSSSSSSLTAVSAMSSTSAVDPSLTRPPEELTLSPKLQLDGSLTMSSSGSLQASPRGLLPGLLPAPADKLTPKGPGQVPTATSALSLELQEVEP
>sp|O14683|P5I11_HUMAN Tumor protein p53-inducible protein 11 OS=Homo sapiens OX=9606 GN=TP53I11 PE=1 SV=2
MIHNYMEHLERTKLHQLSGSDQLESTAHSRIRKERPISLGIFPLPAGDGLLTPDAQKGGETPGSEQWKFQELSQPRSHTSLKVSNSPEPQKAVEQEDELSDVSQGGSKATTPASTANSDVATIPTDTPLKEENEGFVKVTDAPNKSEISKHIEVQVAQETRNVSTGSAENEEKSEVQAIIESTPELDMDKDLSGYKGSSTPTKGIENKAFDRNTESLFEELSSAGSGLIGDVDEGADLLGMGREVENLILENTQLLETKNALNIVKNDLIAKVDELTCEKDVLQGELEAVKQAKLKLEEKNRELEEELRKARAEAEDARQKAKDDDDSDIPTAQRKRFTRVEMARVLMERNQYKERLMELQEAVRWTEMIRASRENPAMQEKKRSSIWQFFSRLFSSSSNTTKKPEPPVNLKYNAPTSHVTPSVK
I want to count how many of each letter is there, so if I have one I count like this
>sp|Q96A73|P33MX_HUMAN Putative monooxygenase p33MONOX OS=Homo sapiens OX=9606 GN=KIAA1191 PE=1 SV=1
RNDDDDTSVCLGTRQCSWFAGCTNRTWNSSAVPLIGLPNTQDYKWVDRNSGLTWSGNDTCLYSCQNQTKGLLYQLFRNLFCSYGLTEAHGKWRCADASITNDKGHDGHRTPTWWLTGSNLTLSVNNSGLFFLCGNGVYKGFPPKWSGRCGLGYLVPSLTRYLTLNASQITNLRSFIHKVTPHR
cat input.txt | grep -v ">" | fold -w1 | sort | uniq -c
6 A
9 C
10 D
1 E
7 F
18 G
5 H
4 I
7 K
21 L
15 N
7 P
6 Q
11 R
16 S
18 T
7 V
8 W
7 Y
however, I want to calculate for all in a better way and more efficient especially when the data is huge
回答1:
Counting characters in strings can easily be done with awk. To do this, you make use of the function gsub
:
gsub(ere, repl[, in])
Behave likesub
(see below), except that it shall replace all occurrences of the regular expression (like the ed utility global substitute) in$0
or in thein
argument when specified.
sub(ere, repl[, in ])
Substitute the stringrepl
in place of the first instance of the extended regular expressionERE
in string in and return the number of substitutions. <snip> Ifin
is omitted, awk shall use the current record ($0
) in its place.source: Awk Posix Standard
The following two functions perform the counting in this way:
function countCharacters(str) {
while(str != "") { c=substr(str,1,1); a[toupper[c]]+=gsub(c,"",str) }
}
or if there might appear a lot of equal consecutive characters, the following solution might shave off a couple of seconds.
function countCharacters2(str) {
n=length(str)
while(str != "") { c=substr(str,1,1); gsub(c"+","",str);
m=length(str); a[toupper[c]]+=n-m; n=m
}
}
Below you find 4 implementations based on the first function. The first two run on a standard awk, the latter two on an optimized version for fasta-files:
1. Read sequence and processes it line by line:
awk '!/^>/{s=$0; while(s!="") { c=substr(s,1,1); a[c]+=gsub(c,"",s) } }
END {for(c in a) print c,a[c]}' file
2. concatenate all sequences and process it in the end:
awk '!/^>/{s=s $0 }
END {while(s!="") { c=substr(s,1,1); a[c]+=gsub(c,"",s) }
for(c in a) print c,a[c]}' file
3. Same as 1 but use bioawk
:
bioawk -c fastx '{while ($seq!=""){ c=substr($seq,1,1);a[c]+=gsub(c,"",$seq) } }
END{ for(c in a) print c,a[c] }' file
4. Same as 2 but use bioawk
:
bioawk -c fastx '{s=s $seq}
END { while(s!="") { c=substr(s,1,1); a[c]+=gsub(c,"",s) }
for(c in a) print c,a[c]}' file
Here are some timing results based on this fasta-file
OP : grep,sort,uniq : 47.548 s
EdMorton 1 : awk : 39.992 s
EdMorton 2 : awk,sort,uniq : 53.965 s
kvantour 1 : awk : 18.661 s
kvantour 2 : awk : 9.309 s
kvantour 3 : bioawk : 1.838 s
kvantour 4 : bioawk : 1.838 s
karafka : awk : 38.139 s
stack0114106 1: perl : 22.754 s
stack0114106 2: perl : 13.648 s
stack0114106 3: perl (zdim) : 7.759 s
Note: BioAwk is based on Brian Kernighan's awk which is documented in "The AWK Programming Language", by Al Aho, Brian Kernighan, and Peter Weinberger (Addison-Wesley, 1988, ISBN 0-201-07981-X) . I'm not sure if this version is compatible with POSIX.
回答2:
With any awk in any shell on any UNIX box:
$ cat tst.awk
!/^>/ {
for (i=1; i<=length($0); i++) {
cnt[substr($0,i,1)]++
}
}
END {
for (char in cnt) {
print char, cnt[char]
}
}
$ awk -f tst.awk file
A 107
N 67
P 107
C 41
Q 88
D 102
E 132
R 101
F 65
S 168
G 140
T 115
H 52
I 84
V 101
W 27
K 114
Y 30
L 174
M 39
or if you prefer:
$ awk -v ORS= '!/^>/{gsub(/./,"&\n"); print}' file | sort | uniq -c
107 A
41 C
102 D
132 E
65 F
140 G
52 H
84 I
114 K
174 L
39 M
67 N
107 P
88 Q
101 R
168 S
115 T
101 V
27 W
30 Y
回答3:
Try this Perl solution for better performance.
$ perl -lne '
if( ! /^>/ ) { while(/./g) { $kv{$&}++} }
END { for ( sort keys %kv) { print "$_ $kv{$_}" }}
' learner.txt
A 107
C 41
D 102
E 132
F 65
G 140
H 52
I 84
K 114
L 174
M 39
N 67
P 107
Q 88
R 101
S 168
T 115
V 101
W 27
Y 30
$
One more solution using Perl, optimized for performance.
$ time perl -lne '
if( ! /^>/ ) { for($i=0;$i<length($_);$i++)
{ $x=substr($_,$i,1); $kv{$x}++ } }
END { for ( sort keys %kv) { print "$_ $kv{$_}" }}
' chrY.fa
A 2994088
C 1876822
G 1889305
N 30812232
T 3002884
a 4892104
c 3408967
g 3397589
n 140
t 4953284
real 0m15.922s
user 0m15.750s
sys 0m0.108s
$
Edit with further performance optimizations
All timings reported below are averages over 3-5 runs on a desktop, done at around the same time but swapped around to avoid pronounced cacheing effects.
Changing the C-style for
loop to for my $i (0..length($_))
speeds the second solution from 9.2 seconds to 6.8 seconds.
Then, also removing a scalar ($x
) at each operation, with
if (not /^>/) { for $i (0..length($_)) { ++$kv{ substr($_,$i,1) } } }
speeds this up to 5.3 seconds.
Further reducing variable use, by copying $_
and thus freeing up the loop to use $_
if (not /^>/) { $l=$_; ++$kv{ substr($l,$_,1) } for 0..length($l) }
only helps a little, running at 5.2 seconds.
This compares with the awk
solution, given as kvantour 2
in nice comparisons in kvantour answer, at 6.5 seconds (on this system).
Of course none of this can be compared to the optimized bioawk
(C-code?) program. For that we'd need to write this in C (which is not very hard using Inline C
).
Note that removing a sub call (to substr
) for every character by using
if (not /^>/) { ++$kv{$_} for split //; }
results in "only" a 6.4 seconds average, not as good as the above tweaks; this was a surprise.
These times are on a desktop with v5.16. On v5.24, on the same machine, the best-case (substr
with no extra variables in the loop) time is 4.8 seconds while the one without the substr
(but with split
) is 5.8 seconds. It's nice to see that newer versions of Perl perform better, at least in these cases.
For reference and easy timing by others, complete code for the best run
time perl -lne'
if (not /^>/) { $l=$_; ++$kv{ substr($l,$_,1) } for 0..length($l) }
END { for ( sort keys %kv) { print "$_ $kv{$_}" }}
' chrY.fa
回答4:
not sure how much faster this would be but if you try please post your timings
$ awk '!/^>/ {n=split($0,a,""); for(i=1;i<=n;i++) c[a[i]]++}
END {for(k in c) print k,c[k]}' file | sort
A 6
C 9
D 10
E 1
F 7
G 18
H 5
I 4
K 7
L 21
N 15
P 7
Q 6
R 11
S 16
T 18
V 7
W 8
Y 7
this reports counts for the file, not line by line. As noted below, not all awk
's support empty string split.
Here are the timings of the three approaches:
$ time grep -v ">" filey | fold -w1 | sort | uniq -c >/dev/null
real 0m11.470s
user 0m11.746s
sys 0m0.260s
$ time awk '{n=split($0,a,""); for(i=1;i<=n;i++) c[a[i]++]} END{for(k in c) print k,c[k]}' filey >/dev/null
real 0m7.441s
user 0m7.334s
sys 0m0.060s
$ time awk '{n=length($0); for(i=1;i<=n;i++) c[substr($0,i,1)]++} END{for(k in c) print k,c[k]}' filey >/dev/null
real 0m5.055s
user 0m4.979s
sys 0m0.047s
for the test file
$ wc filey
118098 649539 16828965 filey
it surprised me that substr
is faster than split
. Perhaps due to array allocation.
来源:https://stackoverflow.com/questions/54712459/how-can-i-count-the-frequency-of-letters