awk command to count specific field

Question

[saadleeshehreen]

Hi, I have a file with the following content. Now I want to count how many of them have 1 in the field -n2.

Bacteroides fragilis,0
Bacteroides fragilis,0
Salmonella enterica,1
Salmonella enterica,1
Salmonella enterica,1
Bacteroides fragilis,0
.................................
..................................

I used the following command :

cat f1.txt | awk {'$2 == 1'} | wc -l

But it doesn't give me the answer. Please help!

[MSM55]

I am not able to Understand input file format. You can use following command to count number of 1 in field 2

grep -o ",1" input.txt  | wc -l

Answer 1

[lakhujanivijay]

The weird format of your file (if indeed it is in this way) is out of anyone's understanding. But I ll explain how awk could work here provided a nicely formatted tab separated table

Consider your file (say file.txt) this way, the <tab> and <space> symbols are for representation, your actual file will have whitepspace (tabs and spaces) and corresponding positions shown in the file

Bacteroides<space>fragilis<tab>0
Bacteroides<space>fragilis<tab>0
Salmonella<space>enterica<tab>1
Salmonella<space>enterica<tab>1
Salmonella<space>enterica<tab>1
Bacteroides<space>fragilis<tab>0

Now if you say

awk '$2==1{print}' file.txt | wc -l

It may not work, because, by default the field separator which awk consider here is the first white space it encounters which in this case would be the space Bacteroides <space> fragilis

Hence, you must add a field separator -F

awk -F "\t" '$2==1{print}' file.txt | wc -l

[Friederike]

awk -F "," '$2==1' file.txt | wc -l

Answer 2

[Pierre Lindenbaum]

set the field separator 'F' to 'comma' and increase a value 'N' each time column 2 is '1'. At the end print the value of N.

awk -F, '($2==1){N++;}END{print N;}' file.txt

but I think most people would use

cut -d, -f 2 file.txt | grep -c -w 1

[saadleeshehreen]

Hi, Thanks. I have other related problem. My file like this:

10 Lachnoclostridium sp.   0       0       0       0       1
11 Haemophilus ducreyi     0       0       0       0       1
12 Clostridiales bacterium 0       0       0       0       1
13 Escherichia albertii    0       1       0       0       1

It has 8 fields. I want to just count the lines which value =1 in field 7 and field 8. How can I do that? I used the following, but it's not the exact output.

awk '$4 == 0; $5 == 0; $6 == 0; $7 == 1; $8 ==1' file.txt

[Pierre Lindenbaum]

https://www.gnu.org/software/gawk/manual/html_node/Boolean-Ops.html

[MSM55]

Hi, You can use following command

awk  '$7==1 && $8==1 {print}' input.txt

[Pierre Lindenbaum]

or just awk '$7==1 && $8==1' input.txt

[saadleeshehreen]

Thanks. It works for me. :)

[Pierre Lindenbaum]

If an answer was helpful you should upvote it, if the answer resolved your question you should mark it as accepted.

[cpad0112]

try this as well:

awk '$7 && $8==1'  input.txt

Answer 3

[cpad0112]

Though OP wants solution in awk, here is datamash solution:

Species (organism) wide 0's and 1's count:

$ datamash -s -t "," -g 1,2 count 2 < test.txt | sed 's/,/\t/g'
Bacteroides fragilis    0   3
Salmonella enterica 1   3

Only 0's and 1's count:

$ datamash -s -t "," -g 2 count 2 < test.txt | sed 's/,/\t/g'
0   3
1   3

input (from OP):

 $ cat test.txt 
Bacteroides fragilis,0
Bacteroides fragilis,0
Salmonella enterica,1
Salmonella enterica,1
Salmonella enterica,1
Bacteroides fragilis,0