awk: add column name with filename in loop2019 Community Moderator Electionchange column with awk scriptingSeparating names separated by “| character” to individual pairsawk change one column with anotherawk Command Loopstat(FILENAME) in awkApplying a mathematical function to every line in a large fileUsing 'awk' to print a placeholder in several blocks of dataPrint column with AWKget average for all numbers with irregular columnsAWK doesn't add new column
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awk: add column name with filename in loop
2019 Community Moderator Electionchange column with awk scriptingSeparating names separated by “| character” to individual pairsawk change one column with anotherawk Command Loopstat(FILENAME) in awkApplying a mathematical function to every line in a large fileUsing 'awk' to print a placeholder in several blocks of dataPrint column with AWKget average for all numbers with irregular columnsAWK doesn't add new column
I have a question to add a column title during a loop with awk:
I currently have this code, which calculates the mean of each line for each csv file :
for i in *.csv; do
awk -F"," 'NR==1 next
T=0
for(N=1; N<=NF; N++) T+=$N;
T/=NF
print T ' $i > Mean$i
done
So I get files MeanName1.csv, MeanName2.csv, ... containing:
2.5 3.5 ...
4.2 3.4
1.2 3.9
1.2 3.7
I would like to get a single file "Mean" containing:
Name1 Name2 ...
2.5 3.5
4.2 3.4
1.2 3.9
1.2 3.7
I failed to use print FILENAME in the loop.
Do you have any suggestions ?
Thank you
awk
asked 2 hours ago
MLdishMLdish
84
New contributor
MLdish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I have a question to add a column title during a loop with awk:
I currently have this code, which calculates the mean of each line for each csv file :
for i in *.csv; do
awk -F"," 'NR==1 next
T=0
for(N=1; N<=NF; N++) T+=$N;
T/=NF
print T ' $i > Mean$i
done
So I get files MeanName1.csv, MeanName2.csv, ... containing:
2.5 3.5 ...
4.2 3.4
1.2 3.9
1.2 3.7
I would like to get a single file "Mean" containing:
Name1 Name2 ...
2.5 3.5
4.2 3.4
1.2 3.9
1.2 3.7
I failed to use print FILENAME in the loop.
Do you have any suggestions ?
Thank you
awk
asked 2 hours ago
MLdishMLdish
84
New contributor
MLdish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The way you have written the data in the question makes it look like you have several columns of data in the same file. Can we assume that you only have a single column of data per file, as indicated by your actual code?
– Kusalananda
1 hour ago
Correct, I have only a single column of data in each Mean File.
– MLdish
1 hour ago
add a comment |
I have a question to add a column title during a loop with awk:
I currently have this code, which calculates the mean of each line for each csv file :
for i in *.csv; do
awk -F"," 'NR==1 next
T=0
for(N=1; N<=NF; N++) T+=$N;
T/=NF
print T ' $i > Mean$i
done
So I get files MeanName1.csv, MeanName2.csv, ... containing:
2.5 3.5 ...
4.2 3.4
1.2 3.9
1.2 3.7
I would like to get a single file "Mean" containing:
Name1 Name2 ...
2.5 3.5
4.2 3.4
1.2 3.9
1.2 3.7
I failed to use print FILENAME in the loop.
Do you have any suggestions ?
Thank you
awk
asked 2 hours ago
MLdishMLdish
84
New contributor
MLdish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a question to add a column title during a loop with awk:
I currently have this code, which calculates the mean of each line for each csv file :
for i in *.csv; do
awk -F"," 'NR==1 next
T=0
for(N=1; N<=NF; N++) T+=$N;
T/=NF
print T ' $i > Mean$i
done
So I get files MeanName1.csv, MeanName2.csv, ... containing:
2.5 3.5 ...
4.2 3.4
1.2 3.9
1.2 3.7
I would like to get a single file "Mean" containing:
Name1 Name2 ...
2.5 3.5
4.2 3.4
1.2 3.9
1.2 3.7
I failed to use print FILENAME in the loop.
Do you have any suggestions ?
Thank you
awk
awk
asked 2 hours ago
MLdishMLdish
84
New contributor
MLdish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
MLdishMLdish
84
New contributor
MLdish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
MLdish
asked 2 hours ago
MLdishMLdish
84
New contributor
MLdish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
MLdishMLdish
84
asked 2 hours ago
MLdishMLdish
84
84
New contributor
MLdish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
MLdish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
MLdish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The way you have written the data in the question makes it look like you have several columns of data in the same file. Can we assume that you only have a single column of data per file, as indicated by your actual code?
– Kusalananda
1 hour ago
Correct, I have only a single column of data in each Mean File.
– MLdish
1 hour ago
add a comment |
The way you have written the data in the question makes it look like you have several columns of data in the same file. Can we assume that you only have a single column of data per file, as indicated by your actual code?
– Kusalananda
1 hour ago
Correct, I have only a single column of data in each Mean File.
– MLdish
1 hour ago
The way you have written the data in the question makes it look like you have several columns of data in the same file. Can we assume that you only have a single column of data per file, as indicated by your actual code?
– Kusalananda
1 hour ago
The way you have written the data in the question makes it look like you have several columns of data in the same file. Can we assume that you only have a single column of data per file, as indicated by your actual code?
– Kusalananda
1 hour ago
Correct, I have only a single column of data in each Mean File.
– MLdish
1 hour ago
Correct, I have only a single column of data in each Mean File.
– MLdish
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
Joining the output files is easiest done with paste, but I don't think it knows how to add the file names as headers. But we can do that manually to the intermediate output files within the loop, so:
for i in *.csv; do
out="Mean$i"
echo "$i%.csv" > "$out"
awk -F"," 'NR==1 next
T=0
for(N=1; N<=NF; N++) T+=$N;
T/=NF
print T ' $i >> "$out"
done
paste Mean*.csv > All.csv
$i%.csv is just the value of $i with the suffix .csv removed, so for foo.csv and bar.csv, that gives an output like this:
bar foo
4.33333 2.5
3.8 3.5
I note you're also skipping the first line of each input line (NR==1 next ). If they contain useful headers, you could of course use those too by changing that to NR==1 print; next and then removing the echo I added above.
answered 1 hour ago
ilkkachuilkkachu
61k1098174
Exactly what I needed, thanks! (I don't need header of each *.csv files, so I keep echo)
– MLdish
1 hour ago
add a comment |
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votes
Joining the output files is easiest done with paste, but I don't think it knows how to add the file names as headers. But we can do that manually to the intermediate output files within the loop, so:
for i in *.csv; do
out="Mean$i"
echo "$i%.csv" > "$out"
awk -F"," 'NR==1 next
T=0
for(N=1; N<=NF; N++) T+=$N;
T/=NF
print T ' $i >> "$out"
done
paste Mean*.csv > All.csv
$i%.csv is just the value of $i with the suffix .csv removed, so for foo.csv and bar.csv, that gives an output like this:
bar foo
4.33333 2.5
3.8 3.5
I note you're also skipping the first line of each input line (NR==1 next ). If they contain useful headers, you could of course use those too by changing that to NR==1 print; next and then removing the echo I added above.
answered 1 hour ago
ilkkachuilkkachu
61k1098174
Exactly what I needed, thanks! (I don't need header of each *.csv files, so I keep echo)
– MLdish
1 hour ago
add a comment |
Joining the output files is easiest done with paste, but I don't think it knows how to add the file names as headers. But we can do that manually to the intermediate output files within the loop, so:
for i in *.csv; do
out="Mean$i"
echo "$i%.csv" > "$out"
awk -F"," 'NR==1 next
T=0
for(N=1; N<=NF; N++) T+=$N;
T/=NF
print T ' $i >> "$out"
done
paste Mean*.csv > All.csv
$i%.csv is just the value of $i with the suffix .csv removed, so for foo.csv and bar.csv, that gives an output like this:
bar foo
4.33333 2.5
3.8 3.5
I note you're also skipping the first line of each input line (NR==1 next ). If they contain useful headers, you could of course use those too by changing that to NR==1 print; next and then removing the echo I added above.
answered 1 hour ago
ilkkachuilkkachu
61k1098174
Exactly what I needed, thanks! (I don't need header of each *.csv files, so I keep echo)
– MLdish
1 hour ago
add a comment |
Joining the output files is easiest done with paste, but I don't think it knows how to add the file names as headers. But we can do that manually to the intermediate output files within the loop, so:
for i in *.csv; do
out="Mean$i"
echo "$i%.csv" > "$out"
awk -F"," 'NR==1 next
T=0
for(N=1; N<=NF; N++) T+=$N;
T/=NF
print T ' $i >> "$out"
done
paste Mean*.csv > All.csv
$i%.csv is just the value of $i with the suffix .csv removed, so for foo.csv and bar.csv, that gives an output like this:
bar foo
4.33333 2.5
3.8 3.5
I note you're also skipping the first line of each input line (NR==1 next ). If they contain useful headers, you could of course use those too by changing that to NR==1 print; next and then removing the echo I added above.
answered 1 hour ago
ilkkachuilkkachu
61k1098174
Joining the output files is easiest done with paste, but I don't think it knows how to add the file names as headers. But we can do that manually to the intermediate output files within the loop, so:
for i in *.csv; do
out="Mean$i"
echo "$i%.csv" > "$out"
awk -F"," 'NR==1 next
T=0
for(N=1; N<=NF; N++) T+=$N;
T/=NF
print T ' $i >> "$out"
done
paste Mean*.csv > All.csv
$i%.csv is just the value of $i with the suffix .csv removed, so for foo.csv and bar.csv, that gives an output like this:
bar foo
4.33333 2.5
3.8 3.5
I note you're also skipping the first line of each input line (NR==1 next ). If they contain useful headers, you could of course use those too by changing that to NR==1 print; next and then removing the echo I added above.
answered 1 hour ago
ilkkachuilkkachu
61k1098174
edited 1 hour ago
answered 1 hour ago
ilkkachuilkkachu
61k1098174
answered 1 hour ago
ilkkachuilkkachu
61k1098174
answered 1 hour ago
ilkkachuilkkachu
61k1098174
61k1098174
Exactly what I needed, thanks! (I don't need header of each *.csv files, so I keep echo)
– MLdish
1 hour ago
add a comment |
Exactly what I needed, thanks! (I don't need header of each *.csv files, so I keep echo)
– MLdish
1 hour ago
Exactly what I needed, thanks! (I don't need header of each *.csv files, so I keep echo)
– MLdish
1 hour ago
Exactly what I needed, thanks! (I don't need header of each *.csv files, so I keep echo)
– MLdish
1 hour ago
add a comment |
MLdish is a new contributor. Be nice, and check out our Code of Conduct.
MLdish is a new contributor. Be nice, and check out our Code of Conduct.
MLdish is a new contributor. Be nice, and check out our Code of Conduct.
MLdish is a new contributor. Be nice, and check out our Code of Conduct.
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-awk
The way you have written the data in the question makes it look like you have several columns of data in the same file. Can we assume that you only have a single column of data per file, as indicated by your actual code?
– Kusalananda
1 hour ago
Correct, I have only a single column of data in each Mean File.
– MLdish
1 hour ago