Is there a way to get the date range from a csv file?

Question

I want to find the date range for each of the 4 CSV Files in a directory. The date column is the last column in each csv and is of the form 1/25/2012 7:20:55 PM.

The date field is always the last column that is the 24th column. Only the records in each csv change. And the date is ordered.

Is there a way that I can do it for each of the files and get the final range, assuming that the dates are ordered?

So if my first record is 1/25/2012 7:20:55 PM and last record is 11/7/2016 2:36:20 PM.

I want the output date range as 1/25/2012 - 11/7/2016. But I want to club all the 4 files date range.

Example input (some columns omitted for brevity):

• First file:

  col1,col2,col3,...,col23,col24_time
  value1,value2,value3,...,value23,1/25/2012 7:20 PM
  value1,value2,value3,...,value23,1/26/2012 10:57 AM
  value1,value2,value3,...,value23,1/26/2012 2:20 PM
  value1,value2,value3,...,value23,1/30/2012 11:55 AM
  value1,value2,value3,...,value23,1/30/2012 3:17 PM
  value1,value2,value3,...,value23,1/30/2012 5:36 PM
  value1,value2,value3,...,value23,1/30/2012 8:16 PM
  ...
  value1,value2,value3,...,value23,4/11/2012 11:45 AM
  value1,value2,value3,...,value23,4/11/2012 2:23 PM

• Intermediate files

• Last file:

  value1,value2,value3,...,value23,3/11/2015 4:45 AM
  value1,value2,value3,...,value23,3/11/2015 8:40 AM
  ...
  value1,value2,value3,...,value23,11/7/2016 2:36 PM

I have close to 5-10K records in each file. The dates are ordered within the files. And each file has an header for each column.

The output from the head -n7 Files/file1.csv | cut -d, -f24 command is:

"col24_time"
"2012-01-01 00:30:26"
"0"
"2012-01-01 02:00:37"
"0"
"0"
"https://external.xx.fbcdn.net/safe_image.php?" <<-- previous column record?

Answer 1

I’m still not sure that I understand the question. But here is some code that produces the desired output from the specified input, and it’s much shorter than the other answers:

datetime1=$(head -n1 file1.csv | cut -d, -f24)
datetime4=$(tail -n1 file4.csv | cut -d, -f24)
printf '%s - %s\n' "${datetime1%% *}" "${datetime4%% *}"

This takes the first line from the first file, and the last line from the fourth and final file, and extracts the 24th field (based on , as a delimiter) of each.  These are date-time strings; specifically, 1/25/2012 7:20 AM and 11/7/2016 2:36 PM.  It then prints the first “word” of each, by stripping off the first space and everything that comes after it.  These are the dates, as desired.

Here’s a quasi-one-liner equivalent.  I spread it across three lines for readability, but it’s logically one long command.

printf '%s - %s\n' \
        "$(head -n1 file1.csv | cut -d, -f24 | cut -d' ' -f1)" \
        "$(tail -n1 file4.csv | cut -d, -f24 | cut -d' ' -f1)"

Here, since we aren’t using any variables, we can’t use parameter expansion, so I extracted the first “word” of the 24th fields with a second cut.

Answer 2

If your data is, as you say, already date sequenced and consistently structured then you can use sed to work on your specific lines:

sed -E -n "2 {s/.*,([^ ]*).*/\1 - /;h}; $ {s/.*,([^ ]*).*/\1/;H;x;s/\n//;p}" file

Output for your first file is

1/26/2012 - 4/11/2012

To do all the files together just cat them (assumed to be named in date order and piped in the correct sequence) as

cat file* | sed ...
1/26/2012 - 11/7/2016

Walkthrough

Set sed default as -not to print

sed -E -n "

Grab the 2nd line and collect the part of the line you want ([^ ]+) as a capture group and compose your output in pattern space as your capture and your separator \1 -

           2 {s/.*,([^ ]+).*/\1 - /;

Push that into hold space (h clears out anything that was there before)

                                h}; 

On the la$t line grab the part of the line you want again in pattern space

                                    $ {s/.*,([^ ]+).*/\1/;

Append the new pattern space to the Hold space with a \newline between the old and new content (H adds the \newline) and then exchange the entire contents of hold space with pattern space

                                                H;x;

With your composed output now in pattern space, just remove the unwanted \newline and print

                                                    s/\n//;p}" file

Answer 3

The following awk program will work (let's call it timerange.awk). It is designed so that you don't have to provide the files in any specific order but can simply use it on *.csv - otherwise, you would only need to provide the first and last file in the directory anyway, as you state that timestamps are ordered.

#!/usr/bin/awk -f
For every line of the files (after the first, which contains headers)
FNR>1{
    # Break the time stamp field into its individual components and reassemble
    # in a way that 'mktime' understands, to generate an epoch-based timestamp
    # for "later/earlier than"-type comparisons.
    split($NF,a,/[ /:]/);
    if (a[6]=="AM" && a[4]==12) a[4]=0;
    if (a[6]=="PM") a[4]=a[4]+12;
    tst=a[3]" " a[1] " " a[2] " " a[4] " " a[5] " 00";
    curr_ts=mktime(tst);
# If we are on the first "data" row of the first file, initialize start and end
# date
if (NR==2)
{
    end=start=$NF;
    end_ts=start_ts=curr_ts;
}
# On all later lines, check if the timestamps associated with the "start"
# and "end" time specifications are later resp. earlier than that of the
# current line. If so, update "start" and "end" specifications.
else
{
    if (curr_ts>end_ts) {end_ts=curr_ts; end=$NF};
    if (curr_ts<start_ts) {start_ts=curr_ts; start=$NF};
}

}
After the last file was processed: Output the human-readable range
END{print start " - " end}

You can call it as

awk -F, -f timerange.awk file1.csv file2.csv ...

or simply

awk -F, -f timerange.awk *.csv

as the ordering of the files is not relevant; it will automatically look for the global "first" and "last" entry.

Output for the example input (in its current form - I have assumed that contrary to your first statement, the timestamp does not include seconds):

1/25/2012 10:57 AM - 11/7/2016 2:36 PM

Update

If you want to completely ignore the time of day, the program can be abridged:

#!/usr/bin/awk -f
For every line of the files (after the first, which contains headers)
FNR>1{
    split($NF,a,/[ /:]/);
    tst=a[3]" " a[1] " " a[2] " 00 00 00"
    curr_ts=mktime(tst);
sub(/[[:space:]]+.* [AP]M$/,"",$NF);

if (NR==2)
{
    end=start=$NF;
    end_ts=start_ts=curr_ts;
}
else
{
    if (curr_ts>end_ts) {end_ts=curr_ts; end=$NF};
    if (curr_ts<start_ts) {start_ts=curr_ts; start=$NF};
}

}
END{print start " - " end}

Answer 4

If you like Unix pipes, you can do it this way:

# standalone example: this converts from a 2-colum, 1-line "csv" to unixtime, 
#  and converts back to readable date
echo "2,1/25/2012  7:20:55 PM" \
| perl -aF, -MDate::Parse -E "say Date::Parse::str2time(\$F[1])" - \
| xargs -i date "+%D " -d@{}
result
01/25/12

This relies on the old but non-core Perl module Date::Parse, which you'd need to install first if you don't have it already.
Perhaps install with cpan Date::Parse, or with the zero-configuration client cpanm Date::Parse.

So for your example you could try this to get youngest and oldest on two single lines

perl -aF, -MDate::Parse -E "say Date::Parse::str2time(\$F[5])" *.csv \
| sort \
| sed -e 1b -e '$!d'  \
| xargs -i date "+%D " -d@{}
result
01/25/12 
11/07/16

The sed line is from this post on this site.