GREP values from a column in txt file - regex

Question

I have a txt file with 1200 entries in this way (iPerf output by the way)
1 [ 4] 0.0- 1.0 sec 10.6 MBytes 89.1 Mbits/sec
2 [ 4] 1.0- 2.0 sec 13.5 MBytes 113 Mbits/sec
3 [ 4] 2.0- 3.0 sec 9.50 MBytes 79.7 Mbits/sec
4 [ 4] 3.0- 4.0 sec 9.00 MBytes 75.5 Mbits/sec
How can I get ONLY the second values expressed in Mbits/sec using grep ?
Output example:
89.1
113
79.7
75.5

Answer 1

awk '{print $9}' your-file.txt
will do it for you. For example:
$ cat ~/test.txt
1 [ 4] 0.0- 1.0 sec 10.6 MBytes 89.1 Mbits/sec
2 [ 4] 1.0- 2.0 sec 13.5 MBytes 113 Mbits/sec
3 [ 4] 2.0- 3.0 sec 9.50 MBytes 79.7 Mbits/sec
4 [ 4] 3.0- 4.0 sec 9.00 MBytes 75.5 Mbits/sec
$ awk '{print $9}' ~/test.txt
89.1
113
79.7
75.5
Another way to tackle this is:
awk -F 'MBytes' '{print $2}' test.txt | awk -F 'Mbits' '{print $1}' | tr -d " "
In the above method we are:
Splitting each line by MBytes.
That gives us 2 parts: $1 is everything before MBytes. $2 is everything after MBytes
We choose everything after MBytes and split it further by Mbits
That gives us two parts again and we choose everything before Mbits
If there is white space before and after the numbers, we use tr to remove white space
So we get
$ cat test.txt
1 [ 4] 0.0- 1.0 sec 10.6 MBytes 89.1 Mbits/sec
2 [ 4] 1.0- 2.0 sec 13.5 MBytes 113 Mbits/sec
3 [ 4] 2.0- 3.0 sec 9.50 MBytes 79.7 Mbits/sec
4 [ 4] 3.0- 4.0 sec 9.00 MBytes 75.5 Mbits/sec
awk -F 'MBytes' '{print $2}' test.txt | awk -F 'Mbits' '{print $1}' | tr -d " "
Result:
89.1
113
79.7
75.5

Answer 2

if your data is fixed length format you can always use cut
cut -c38-41 data
if you know that the values are 4 chars wide.