AWK print the value from column $3 in column $4 based on the last duplicate from column $1
I have an input file.
30.01.2025 47288.39 47288.39
30.01.2025 -1.00 47287.39
31.01.2025 50.00 47337.39
31.01.2025 -1600.00 45737.39
31.01.2025 1880.00 47617.39
31.01.2025 18.21 47635.60
31.01.2025 -3.46 47632.14
03.02.2025 -51.00 47581.14
03.02.2025 -10.00 47571.14
04.02.2025 -1600.00 45971.14
04.02.2025 1609.00 47580.14
05.02.2025 -800.00 46780.14
05.02.2025 -320.00 46460.14
05.02.2025 51.00 46511.14
05.02.2025 800.00 47311.14
05.02.2025 11.00 47322.14
I want to get an output file.
30.01.2025 47288.39 47288.39
30.01.2025 -1.00 47287.39 47287.39
31.01.2025 50.00 47337.39
31.01.2025 -1600.00 45737.39
31.01.2025 1880.00 47617.39
31.01.2025 18.21 47635.60
31.01.2025 -3.46 47632.14 47632.14
03.02.2025 -51.00 47581.14
03.02.2025 -10.00 47571.14 47571.14
04.02.2025 -1600.00 45971.14
04.02.2025 1609.00 47580.14 47580.14
05.02.2025 -800.00 46780.14
05.02.2025 -320.00 46460.14
05.02.2025 51.00 46511.14
05.02.2025 800.00 47311.14
05.02.2025 11.00 47322.14 47322.14
I tried.
awk '{if(!seen[$1]++){print $0}}' File_1
Solution
You want to duplicate $3 when the current record is that last one for that day.
I find it easier to reverse the lines, do something for the first "whatever", and then re-reverse the lines:
tac File_1 | awk '$1 != prev {prev = $1; $0 = $0 FS $3} 1' | tac
Ordinarily I would write $4 = $3 but that will destroy the spacing for the updated lines.
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