Mixing
Make awk produce error on non-numeric
I have a program that sums a column in a file:
awk -v col=2 '{sum+=$col}END{print sum}' input-file
However, it has a problem: If you give it a file that doesn't have numeric data, (or if one number is missing) it will interpret it as zero.
I want it to produce an error if one of the fields cannot be parsed as a number.
Here's an example input:
bob 1
dave 2
alice 3.5
foo bar
I want it to produce an error because 'bar' is not a number, rather than ignoring the error.
awk numeric-data
asked Jan 16 at 17:18
Nick ODellNick ODell
1,0912920
add a comment |
I have a program that sums a column in a file:
awk -v col=2 '{sum+=$col}END{print sum}' input-file
However, it has a problem: If you give it a file that doesn't have numeric data, (or if one number is missing) it will interpret it as zero.
I want it to produce an error if one of the fields cannot be parsed as a number.
Here's an example input:
bob 1
dave 2
alice 3.5
foo bar
I want it to produce an error because 'bar' is not a number, rather than ignoring the error.
awk numeric-data
asked Jan 16 at 17:18
Nick ODellNick ODell
1,0912920
2
If not a dupe, at least strongly related: Can I determine type of an awk variable?
– Kusalananda
Jan 16 at 17:30
by "produce an error", do you mean stop altogether, or skip the line, and/or emit a message?
– Jeff Schaller
Jan 16 at 17:39
Stop altogether and emit a message.
– Nick ODell
Jan 16 at 17:41
@Kusalananda Thanks, that was really helpful.
– Nick ODell
Jan 16 at 17:54
add a comment |
I have a program that sums a column in a file:
awk -v col=2 '{sum+=$col}END{print sum}' input-file
However, it has a problem: If you give it a file that doesn't have numeric data, (or if one number is missing) it will interpret it as zero.
I want it to produce an error if one of the fields cannot be parsed as a number.
Here's an example input:
bob 1
dave 2
alice 3.5
foo bar
I want it to produce an error because 'bar' is not a number, rather than ignoring the error.
awk numeric-data
asked Jan 16 at 17:18
Nick ODellNick ODell
1,0912920
I have a program that sums a column in a file:
awk -v col=2 '{sum+=$col}END{print sum}' input-file
However, it has a problem: If you give it a file that doesn't have numeric data, (or if one number is missing) it will interpret it as zero.
I want it to produce an error if one of the fields cannot be parsed as a number.
Here's an example input:
bob 1
dave 2
alice 3.5
foo bar
I want it to produce an error because 'bar' is not a number, rather than ignoring the error.
awk numeric-data
awk numeric-data
asked Jan 16 at 17:18
Nick ODellNick ODell
1,0912920
asked Jan 16 at 17:18
Nick ODellNick ODell
1,0912920
asked Jan 16 at 17:18
Nick ODellNick ODell
1,0912920
asked Jan 16 at 17:18
Nick ODellNick ODell
1,0912920
asked Jan 16 at 17:18
Nick ODellNick ODell
1,0912920
1,0912920
2
If not a dupe, at least strongly related: Can I determine type of an awk variable?
– Kusalananda
Jan 16 at 17:30
by "produce an error", do you mean stop altogether, or skip the line, and/or emit a message?
– Jeff Schaller
Jan 16 at 17:39
Stop altogether and emit a message.
– Nick ODell
Jan 16 at 17:41
@Kusalananda Thanks, that was really helpful.
– Nick ODell
Jan 16 at 17:54
add a comment |
2
If not a dupe, at least strongly related: Can I determine type of an awk variable?
– Kusalananda
Jan 16 at 17:30
by "produce an error", do you mean stop altogether, or skip the line, and/or emit a message?
– Jeff Schaller
Jan 16 at 17:39
Stop altogether and emit a message.
– Nick ODell
Jan 16 at 17:41
@Kusalananda Thanks, that was really helpful.
– Nick ODell
Jan 16 at 17:54
2
2
If not a dupe, at least strongly related: Can I determine type of an awk variable?
– Kusalananda
Jan 16 at 17:30
If not a dupe, at least strongly related: Can I determine type of an awk variable?
– Kusalananda
Jan 16 at 17:30
by "produce an error", do you mean stop altogether, or skip the line, and/or emit a message?
– Jeff Schaller
Jan 16 at 17:39
by "produce an error", do you mean stop altogether, or skip the line, and/or emit a message?
– Jeff Schaller
Jan 16 at 17:39
Stop altogether and emit a message.
– Nick ODell
Jan 16 at 17:41
Stop altogether and emit a message.
– Nick ODell
Jan 16 at 17:41
@Kusalananda Thanks, that was really helpful.
– Nick ODell
Jan 16 at 17:54
@Kusalananda Thanks, that was really helpful.
– Nick ODell
Jan 16 at 17:54
add a comment |
4 Answers
4
active
oldest
votes
A reasonable way to test would be to compare the field using tests similar to strtod, which is the method that awk uses to convert strings to numbers:
$2 !~ / *[+-]?[[:digit:]]/ { print "NAN: " $2; exit 1; }
The above differs from strtod in that it does not consider INFINITY or NAN to be "numbers". The leading space requirement could be relaxed under awk's default field-splitting behavior -- meaning the fields would never contain leading space:
$2 !~ /[+-]?[[:digit:]]/ { print "NAN: " $2; exit 1; }
A further refinement, thanks to Stéphane's comment and answer here:
$2 !~ /^[+-]?([[:digit:]]*.?[[:digit:]]*([eE][-+]?[[:digit:]]+)?|0[xX][[:xdigit:]]*.?[[:xdigit:]]*([pP][-+]?[[:digit:]]+)?)$/ { print "NAN: " $2; exit 1; }
Broken out for slightly better legibility, that regex is:
/^[+-]?([[:digit:]]*.?[[:digit:]]*([eE][-+]?[[:digit:]]+)?|
0[xX][[:xdigit:]]*.?[[:xdigit:]]*([pP][-+]?[[:digit:]]+)?)$/
... where the intention is to allow a possible leading + or -, then either a floating point number or hexadecimal number. The floating point number has optional leading digits, an option separator (here fixed to be a period .), followed by some number of digits, optionally followed by an exponent. The hex number must start with 0x or 0X, followed by hex digits, a separator, more hex digits, and optionally followed by a "power" (exponent). The entire second field must match one of those formats (as anchored by ^ and $). Omitted here, for the purposes of this question, are the NAN and INFINITY options.
Another option would be to force a numeric conversion, then compare it to zero and then further compare the original input to something that would convert to zero; more specifically, does it start with an optional + or -, then is it followed by zeros, or followed by a period and zeros:
{ number=0 + $2;
if (!number && $2 !~ /^[+-]?(0+)|.0+/)
print "NAN: "$2;
}
answered Jan 16 at 17:46
Jeff SchallerJeff Schaller
42k1156133
1
You'd need to anchor the regexp (with^), otherwise, it's just a test whether the second field contains a decimal digit. You'd also need to take care of numbers like.123. There's also whether you want to honour user's decimal radix (comma vs period). The last one would let strings like.,+,+.
– Stéphane Chazelas
Jan 16 at 21:49
Great points, Stéphane; thank you. I will correct the simpler aspects shortly.
– Jeff Schaller
Jan 17 at 0:43
notice that when converting strings to numbers awk will ignore any trailing junk (see my answer) and/^[+-]?0*.?0*$/is far from being a thorough pattern matching strings that will be completely parsed to0bystrtod(what about0e13,.0e-0,0x0,0x.0p+13, etc?)
– Uncle Billy
Jan 17 at 1:58
Point well taken, @UncleBilly, as also noted by Stéphane above. I'm working on the "zero" regex right now...
– Jeff Schaller
Jan 17 at 2:00
add a comment |
I ended up with this:
awk -v col=$col '
typeof($col) != "strnum" {
print "Error on line " NR ": " $col " is not numeric"
noprint=1
exit 1
}
{
sum+=$col
}
END {
if(!noprint)
print sum
}' $file
This uses typeof, which is a GNU awk extension. typeof($col) returns 'strnum' if $col is a valid number, and 'string' or 'unassigned' if it is not.
See Can I determine type of an awk variable?
answered Jan 16 at 17:51
Nick ODellNick ODell
1,0912920
add a comment |
awk -v col=2 '
$col+0==0 && $col!~/^[+-]?0/ { print "bad number " $col > "/dev/stderr" }
{sum+=$col}
END{print sum}' input-file
It's up to you to complicate it if you want it to also handle .0 or .0e+33 as valid representations of 0; notice that awk will ignore trailing junk when converting strings to numbers ("1.4e1e3"+0, "1.4e1.e7"+0 or "14+13"+0 will be all equal to 14).
answered Jan 17 at 1:53
Uncle BillyUncle Billy
5925
add a comment |
If you give it a file that doesn't have numeric data,
$col ~ /[^-.[:digit:]]/ {print "Error, non-numeric :"; print $col; exit 1};
Explanation just use a RegEx to check for the presence of characters which are not digits nor floating point, sign, etc.
(or if one number is missing)
add
|| ($col == "")
or
|| (length($col) == 0)
to the rule.
Or you could use a comparison to NF if it's the last column like in your example.
answered Jan 16 at 17:41
DrYakDrYak
1915
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
A reasonable way to test would be to compare the field using tests similar to strtod, which is the method that awk uses to convert strings to numbers:
$2 !~ / *[+-]?[[:digit:]]/ { print "NAN: " $2; exit 1; }
The above differs from strtod in that it does not consider INFINITY or NAN to be "numbers". The leading space requirement could be relaxed under awk's default field-splitting behavior -- meaning the fields would never contain leading space:
$2 !~ /[+-]?[[:digit:]]/ { print "NAN: " $2; exit 1; }
A further refinement, thanks to Stéphane's comment and answer here:
$2 !~ /^[+-]?([[:digit:]]*.?[[:digit:]]*([eE][-+]?[[:digit:]]+)?|0[xX][[:xdigit:]]*.?[[:xdigit:]]*([pP][-+]?[[:digit:]]+)?)$/ { print "NAN: " $2; exit 1; }
Broken out for slightly better legibility, that regex is:
/^[+-]?([[:digit:]]*.?[[:digit:]]*([eE][-+]?[[:digit:]]+)?|
0[xX][[:xdigit:]]*.?[[:xdigit:]]*([pP][-+]?[[:digit:]]+)?)$/
... where the intention is to allow a possible leading + or -, then either a floating point number or hexadecimal number. The floating point number has optional leading digits, an option separator (here fixed to be a period .), followed by some number of digits, optionally followed by an exponent. The hex number must start with 0x or 0X, followed by hex digits, a separator, more hex digits, and optionally followed by a "power" (exponent). The entire second field must match one of those formats (as anchored by ^ and $). Omitted here, for the purposes of this question, are the NAN and INFINITY options.
Another option would be to force a numeric conversion, then compare it to zero and then further compare the original input to something that would convert to zero; more specifically, does it start with an optional + or -, then is it followed by zeros, or followed by a period and zeros:
{ number=0 + $2;
if (!number && $2 !~ /^[+-]?(0+)|.0+/)
print "NAN: "$2;
}
answered Jan 16 at 17:46
Jeff SchallerJeff Schaller
42k1156133
1
You'd need to anchor the regexp (with^), otherwise, it's just a test whether the second field contains a decimal digit. You'd also need to take care of numbers like.123. There's also whether you want to honour user's decimal radix (comma vs period). The last one would let strings like.,+,+.
– Stéphane Chazelas
Jan 16 at 21:49
Great points, Stéphane; thank you. I will correct the simpler aspects shortly.
– Jeff Schaller
Jan 17 at 0:43
notice that when converting strings to numbers awk will ignore any trailing junk (see my answer) and/^[+-]?0*.?0*$/is far from being a thorough pattern matching strings that will be completely parsed to0bystrtod(what about0e13,.0e-0,0x0,0x.0p+13, etc?)
– Uncle Billy
Jan 17 at 1:58
Point well taken, @UncleBilly, as also noted by Stéphane above. I'm working on the "zero" regex right now...
– Jeff Schaller
Jan 17 at 2:00
add a comment |
A reasonable way to test would be to compare the field using tests similar to strtod, which is the method that awk uses to convert strings to numbers:
$2 !~ / *[+-]?[[:digit:]]/ { print "NAN: " $2; exit 1; }
The above differs from strtod in that it does not consider INFINITY or NAN to be "numbers". The leading space requirement could be relaxed under awk's default field-splitting behavior -- meaning the fields would never contain leading space:
$2 !~ /[+-]?[[:digit:]]/ { print "NAN: " $2; exit 1; }
A further refinement, thanks to Stéphane's comment and answer here:
$2 !~ /^[+-]?([[:digit:]]*.?[[:digit:]]*([eE][-+]?[[:digit:]]+)?|0[xX][[:xdigit:]]*.?[[:xdigit:]]*([pP][-+]?[[:digit:]]+)?)$/ { print "NAN: " $2; exit 1; }
Broken out for slightly better legibility, that regex is:
/^[+-]?([[:digit:]]*.?[[:digit:]]*([eE][-+]?[[:digit:]]+)?|
0[xX][[:xdigit:]]*.?[[:xdigit:]]*([pP][-+]?[[:digit:]]+)?)$/
... where the intention is to allow a possible leading + or -, then either a floating point number or hexadecimal number. The floating point number has optional leading digits, an option separator (here fixed to be a period .), followed by some number of digits, optionally followed by an exponent. The hex number must start with 0x or 0X, followed by hex digits, a separator, more hex digits, and optionally followed by a "power" (exponent). The entire second field must match one of those formats (as anchored by ^ and $). Omitted here, for the purposes of this question, are the NAN and INFINITY options.
Another option would be to force a numeric conversion, then compare it to zero and then further compare the original input to something that would convert to zero; more specifically, does it start with an optional + or -, then is it followed by zeros, or followed by a period and zeros:
{ number=0 + $2;
if (!number && $2 !~ /^[+-]?(0+)|.0+/)
print "NAN: "$2;
}
answered Jan 16 at 17:46
Jeff SchallerJeff Schaller
42k1156133
1
You'd need to anchor the regexp (with^), otherwise, it's just a test whether the second field contains a decimal digit. You'd also need to take care of numbers like.123. There's also whether you want to honour user's decimal radix (comma vs period). The last one would let strings like.,+,+.
– Stéphane Chazelas
Jan 16 at 21:49
Great points, Stéphane; thank you. I will correct the simpler aspects shortly.
– Jeff Schaller
Jan 17 at 0:43
notice that when converting strings to numbers awk will ignore any trailing junk (see my answer) and/^[+-]?0*.?0*$/is far from being a thorough pattern matching strings that will be completely parsed to0bystrtod(what about0e13,.0e-0,0x0,0x.0p+13, etc?)
– Uncle Billy
Jan 17 at 1:58
Point well taken, @UncleBilly, as also noted by Stéphane above. I'm working on the "zero" regex right now...
– Jeff Schaller
Jan 17 at 2:00
add a comment |
A reasonable way to test would be to compare the field using tests similar to strtod, which is the method that awk uses to convert strings to numbers:
$2 !~ / *[+-]?[[:digit:]]/ { print "NAN: " $2; exit 1; }
The above differs from strtod in that it does not consider INFINITY or NAN to be "numbers". The leading space requirement could be relaxed under awk's default field-splitting behavior -- meaning the fields would never contain leading space:
$2 !~ /[+-]?[[:digit:]]/ { print "NAN: " $2; exit 1; }
A further refinement, thanks to Stéphane's comment and answer here:
$2 !~ /^[+-]?([[:digit:]]*.?[[:digit:]]*([eE][-+]?[[:digit:]]+)?|0[xX][[:xdigit:]]*.?[[:xdigit:]]*([pP][-+]?[[:digit:]]+)?)$/ { print "NAN: " $2; exit 1; }
Broken out for slightly better legibility, that regex is:
/^[+-]?([[:digit:]]*.?[[:digit:]]*([eE][-+]?[[:digit:]]+)?|
0[xX][[:xdigit:]]*.?[[:xdigit:]]*([pP][-+]?[[:digit:]]+)?)$/
... where the intention is to allow a possible leading + or -, then either a floating point number or hexadecimal number. The floating point number has optional leading digits, an option separator (here fixed to be a period .), followed by some number of digits, optionally followed by an exponent. The hex number must start with 0x or 0X, followed by hex digits, a separator, more hex digits, and optionally followed by a "power" (exponent). The entire second field must match one of those formats (as anchored by ^ and $). Omitted here, for the purposes of this question, are the NAN and INFINITY options.
Another option would be to force a numeric conversion, then compare it to zero and then further compare the original input to something that would convert to zero; more specifically, does it start with an optional + or -, then is it followed by zeros, or followed by a period and zeros:
{ number=0 + $2;
if (!number && $2 !~ /^[+-]?(0+)|.0+/)
print "NAN: "$2;
}
answered Jan 16 at 17:46
Jeff SchallerJeff Schaller
42k1156133
A reasonable way to test would be to compare the field using tests similar to strtod, which is the method that awk uses to convert strings to numbers:
$2 !~ / *[+-]?[[:digit:]]/ { print "NAN: " $2; exit 1; }
The above differs from strtod in that it does not consider INFINITY or NAN to be "numbers". The leading space requirement could be relaxed under awk's default field-splitting behavior -- meaning the fields would never contain leading space:
$2 !~ /[+-]?[[:digit:]]/ { print "NAN: " $2; exit 1; }
A further refinement, thanks to Stéphane's comment and answer here:
$2 !~ /^[+-]?([[:digit:]]*.?[[:digit:]]*([eE][-+]?[[:digit:]]+)?|0[xX][[:xdigit:]]*.?[[:xdigit:]]*([pP][-+]?[[:digit:]]+)?)$/ { print "NAN: " $2; exit 1; }
Broken out for slightly better legibility, that regex is:
/^[+-]?([[:digit:]]*.?[[:digit:]]*([eE][-+]?[[:digit:]]+)?|
0[xX][[:xdigit:]]*.?[[:xdigit:]]*([pP][-+]?[[:digit:]]+)?)$/
... where the intention is to allow a possible leading + or -, then either a floating point number or hexadecimal number. The floating point number has optional leading digits, an option separator (here fixed to be a period .), followed by some number of digits, optionally followed by an exponent. The hex number must start with 0x or 0X, followed by hex digits, a separator, more hex digits, and optionally followed by a "power" (exponent). The entire second field must match one of those formats (as anchored by ^ and $). Omitted here, for the purposes of this question, are the NAN and INFINITY options.
Another option would be to force a numeric conversion, then compare it to zero and then further compare the original input to something that would convert to zero; more specifically, does it start with an optional + or -, then is it followed by zeros, or followed by a period and zeros:
{ number=0 + $2;
if (!number && $2 !~ /^[+-]?(0+)|.0+/)
print "NAN: "$2;
}
answered Jan 16 at 17:46
Jeff SchallerJeff Schaller
42k1156133
edited Jan 17 at 2:15
answered Jan 16 at 17:46
Jeff SchallerJeff Schaller
42k1156133
answered Jan 16 at 17:46
Jeff SchallerJeff Schaller
42k1156133
answered Jan 16 at 17:46
Jeff SchallerJeff Schaller
42k1156133
42k1156133
1
You'd need to anchor the regexp (with^), otherwise, it's just a test whether the second field contains a decimal digit. You'd also need to take care of numbers like.123. There's also whether you want to honour user's decimal radix (comma vs period). The last one would let strings like.,+,+.
– Stéphane Chazelas
Jan 16 at 21:49
Great points, Stéphane; thank you. I will correct the simpler aspects shortly.
– Jeff Schaller
Jan 17 at 0:43
notice that when converting strings to numbers awk will ignore any trailing junk (see my answer) and/^[+-]?0*.?0*$/is far from being a thorough pattern matching strings that will be completely parsed to0bystrtod(what about0e13,.0e-0,0x0,0x.0p+13, etc?)
– Uncle Billy
Jan 17 at 1:58
Point well taken, @UncleBilly, as also noted by Stéphane above. I'm working on the "zero" regex right now...
– Jeff Schaller
Jan 17 at 2:00
add a comment |
1
You'd need to anchor the regexp (with^), otherwise, it's just a test whether the second field contains a decimal digit. You'd also need to take care of numbers like.123. There's also whether you want to honour user's decimal radix (comma vs period). The last one would let strings like.,+,+.
– Stéphane Chazelas
Jan 16 at 21:49
Great points, Stéphane; thank you. I will correct the simpler aspects shortly.
– Jeff Schaller
Jan 17 at 0:43
notice that when converting strings to numbers awk will ignore any trailing junk (see my answer) and/^[+-]?0*.?0*$/is far from being a thorough pattern matching strings that will be completely parsed to0bystrtod(what about0e13,.0e-0,0x0,0x.0p+13, etc?)
– Uncle Billy
Jan 17 at 1:58
Point well taken, @UncleBilly, as also noted by Stéphane above. I'm working on the "zero" regex right now...
– Jeff Schaller
Jan 17 at 2:00
1
1
You'd need to anchor the regexp (with
^), otherwise, it's just a test whether the second field contains a decimal digit. You'd also need to take care of numbers like .123. There's also whether you want to honour user's decimal radix (comma vs period). The last one would let strings like ., +, +.– Stéphane Chazelas
Jan 16 at 21:49
You'd need to anchor the regexp (with
^), otherwise, it's just a test whether the second field contains a decimal digit. You'd also need to take care of numbers like .123. There's also whether you want to honour user's decimal radix (comma vs period). The last one would let strings like ., +, +.– Stéphane Chazelas
Jan 16 at 21:49
Great points, Stéphane; thank you. I will correct the simpler aspects shortly.
– Jeff Schaller
Jan 17 at 0:43
Great points, Stéphane; thank you. I will correct the simpler aspects shortly.
– Jeff Schaller
Jan 17 at 0:43
notice that when converting strings to numbers awk will ignore any trailing junk (see my answer) and
/^[+-]?0*.?0*$/ is far from being a thorough pattern matching strings that will be completely parsed to 0 by strtod (what about 0e13, .0e-0, 0x0, 0x.0p+13, etc?)– Uncle Billy
Jan 17 at 1:58
notice that when converting strings to numbers awk will ignore any trailing junk (see my answer) and
/^[+-]?0*.?0*$/ is far from being a thorough pattern matching strings that will be completely parsed to 0 by strtod (what about 0e13, .0e-0, 0x0, 0x.0p+13, etc?)– Uncle Billy
Jan 17 at 1:58
Point well taken, @UncleBilly, as also noted by Stéphane above. I'm working on the "zero" regex right now...
– Jeff Schaller
Jan 17 at 2:00
Point well taken, @UncleBilly, as also noted by Stéphane above. I'm working on the "zero" regex right now...
– Jeff Schaller
Jan 17 at 2:00
add a comment |
I ended up with this:
awk -v col=$col '
typeof($col) != "strnum" {
print "Error on line " NR ": " $col " is not numeric"
noprint=1
exit 1
}
{
sum+=$col
}
END {
if(!noprint)
print sum
}' $file
This uses typeof, which is a GNU awk extension. typeof($col) returns 'strnum' if $col is a valid number, and 'string' or 'unassigned' if it is not.
See Can I determine type of an awk variable?
answered Jan 16 at 17:51
Nick ODellNick ODell
1,0912920
add a comment |
I ended up with this:
awk -v col=$col '
typeof($col) != "strnum" {
print "Error on line " NR ": " $col " is not numeric"
noprint=1
exit 1
}
{
sum+=$col
}
END {
if(!noprint)
print sum
}' $file
This uses typeof, which is a GNU awk extension. typeof($col) returns 'strnum' if $col is a valid number, and 'string' or 'unassigned' if it is not.
See Can I determine type of an awk variable?
answered Jan 16 at 17:51
Nick ODellNick ODell
1,0912920
add a comment |
I ended up with this:
awk -v col=$col '
typeof($col) != "strnum" {
print "Error on line " NR ": " $col " is not numeric"
noprint=1
exit 1
}
{
sum+=$col
}
END {
if(!noprint)
print sum
}' $file
This uses typeof, which is a GNU awk extension. typeof($col) returns 'strnum' if $col is a valid number, and 'string' or 'unassigned' if it is not.
See Can I determine type of an awk variable?
answered Jan 16 at 17:51
Nick ODellNick ODell
1,0912920
I ended up with this:
awk -v col=$col '
typeof($col) != "strnum" {
print "Error on line " NR ": " $col " is not numeric"
noprint=1
exit 1
}
{
sum+=$col
}
END {
if(!noprint)
print sum
}' $file
This uses typeof, which is a GNU awk extension. typeof($col) returns 'strnum' if $col is a valid number, and 'string' or 'unassigned' if it is not.
See Can I determine type of an awk variable?
answered Jan 16 at 17:51
Nick ODellNick ODell
1,0912920
answered Jan 16 at 17:51
Nick ODellNick ODell
1,0912920
answered Jan 16 at 17:51
Nick ODellNick ODell
1,0912920
answered Jan 16 at 17:51
Nick ODellNick ODell
1,0912920
1,0912920
add a comment |
add a comment |
awk -v col=2 '
$col+0==0 && $col!~/^[+-]?0/ { print "bad number " $col > "/dev/stderr" }
{sum+=$col}
END{print sum}' input-file
It's up to you to complicate it if you want it to also handle .0 or .0e+33 as valid representations of 0; notice that awk will ignore trailing junk when converting strings to numbers ("1.4e1e3"+0, "1.4e1.e7"+0 or "14+13"+0 will be all equal to 14).
answered Jan 17 at 1:53
Uncle BillyUncle Billy
5925
add a comment |
awk -v col=2 '
$col+0==0 && $col!~/^[+-]?0/ { print "bad number " $col > "/dev/stderr" }
{sum+=$col}
END{print sum}' input-file
It's up to you to complicate it if you want it to also handle .0 or .0e+33 as valid representations of 0; notice that awk will ignore trailing junk when converting strings to numbers ("1.4e1e3"+0, "1.4e1.e7"+0 or "14+13"+0 will be all equal to 14).
answered Jan 17 at 1:53
Uncle BillyUncle Billy
5925
add a comment |
awk -v col=2 '
$col+0==0 && $col!~/^[+-]?0/ { print "bad number " $col > "/dev/stderr" }
{sum+=$col}
END{print sum}' input-file
It's up to you to complicate it if you want it to also handle .0 or .0e+33 as valid representations of 0; notice that awk will ignore trailing junk when converting strings to numbers ("1.4e1e3"+0, "1.4e1.e7"+0 or "14+13"+0 will be all equal to 14).
answered Jan 17 at 1:53
Uncle BillyUncle Billy
5925
awk -v col=2 '
$col+0==0 && $col!~/^[+-]?0/ { print "bad number " $col > "/dev/stderr" }
{sum+=$col}
END{print sum}' input-file
It's up to you to complicate it if you want it to also handle .0 or .0e+33 as valid representations of 0; notice that awk will ignore trailing junk when converting strings to numbers ("1.4e1e3"+0, "1.4e1.e7"+0 or "14+13"+0 will be all equal to 14).
answered Jan 17 at 1:53
Uncle BillyUncle Billy
5925
answered Jan 17 at 1:53
Uncle BillyUncle Billy
5925
answered Jan 17 at 1:53
Uncle BillyUncle Billy
5925
answered Jan 17 at 1:53
Uncle BillyUncle Billy
5925
5925
add a comment |
add a comment |
If you give it a file that doesn't have numeric data,
$col ~ /[^-.[:digit:]]/ {print "Error, non-numeric :"; print $col; exit 1};
Explanation just use a RegEx to check for the presence of characters which are not digits nor floating point, sign, etc.
(or if one number is missing)
add
|| ($col == "")
or
|| (length($col) == 0)
to the rule.
Or you could use a comparison to NF if it's the last column like in your example.
answered Jan 16 at 17:41
DrYakDrYak
1915
add a comment |
If you give it a file that doesn't have numeric data,
$col ~ /[^-.[:digit:]]/ {print "Error, non-numeric :"; print $col; exit 1};
Explanation just use a RegEx to check for the presence of characters which are not digits nor floating point, sign, etc.
(or if one number is missing)
add
|| ($col == "")
or
|| (length($col) == 0)
to the rule.
Or you could use a comparison to NF if it's the last column like in your example.
answered Jan 16 at 17:41
DrYakDrYak
1915
add a comment |
If you give it a file that doesn't have numeric data,
$col ~ /[^-.[:digit:]]/ {print "Error, non-numeric :"; print $col; exit 1};
Explanation just use a RegEx to check for the presence of characters which are not digits nor floating point, sign, etc.
(or if one number is missing)
add
|| ($col == "")
or
|| (length($col) == 0)
to the rule.
Or you could use a comparison to NF if it's the last column like in your example.
answered Jan 16 at 17:41
DrYakDrYak
1915
If you give it a file that doesn't have numeric data,
$col ~ /[^-.[:digit:]]/ {print "Error, non-numeric :"; print $col; exit 1};
Explanation just use a RegEx to check for the presence of characters which are not digits nor floating point, sign, etc.
(or if one number is missing)
add
|| ($col == "")
or
|| (length($col) == 0)
to the rule.
Or you could use a comparison to NF if it's the last column like in your example.
answered Jan 16 at 17:41
DrYakDrYak
1915
answered Jan 16 at 17:41
DrYakDrYak
1915
answered Jan 16 at 17:41
DrYakDrYak
1915
answered Jan 16 at 17:41
DrYakDrYak
1915
1915
add a comment |
add a comment |
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2
If not a dupe, at least strongly related: Can I determine type of an awk variable?
– Kusalananda
Jan 16 at 17:30
by "produce an error", do you mean stop altogether, or skip the line, and/or emit a message?
– Jeff Schaller
Jan 16 at 17:39
Stop altogether and emit a message.
– Nick ODell
Jan 16 at 17:41
@Kusalananda Thanks, that was really helpful.
– Nick ODell
Jan 16 at 17:54