Calculating distances between multiple points along a line

[Adrián Villalba]

Dear all,

I have a graph (line graph) and i want to know he exact at the y-axis of
all the points of the graph. I think of tracing manually a line over the
line-graph. Then trace another line among the basal line of the x-axis (the
zero in the y-axis) and then calculate the distance (in any unit, it does
not matter) between each point of the graph line and its parallel to the
basal line. It would give me the proportionality respect a reference point
to know each value of the graph line.

The fact is that when i try to do it on ImageJ, i can only do it manually
(which means a huge amount of time) do you know if there is any choice/tool
to perform this calculation along the line-graph?


Thank you all in advance.

--

   - Adrián Villalba Felipe.
   https://es.linkedin.com/in/adrianvillalba

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[Herbie]

Good day Adrián,

the question arises where these graphs come from. If they are not from
your own research, what must be expected, the question arises, whether
you are authorized to use or manipulate them.

In general it is not a good idea to extract data from plots.

Regards

Herbie

:::::::::::::::::::::::::::::::::::::::::::::
Am 19.06.17 um 15:29 schrieb Adrián Villalba:

> Dear all,
>
> I have a graph (line graph) and i want to know he exact at the y-axis of
> all the points of the graph. I think of tracing manually a line over the
> line-graph. Then trace another line among the basal line of the x-axis (the
> zero in the y-axis) and then calculate the distance (in any unit, it does
> not matter) between each point of the graph line and its parallel to the
> basal line. It would give me the proportionality respect a reference point
> to know each value of the graph line.
>
> The fact is that when i try to do it on ImageJ, i can only do it manually
> (which means a huge amount of time) do you know if there is any choice/tool
> to perform this calculation along the line-graph?
>
>
> Thank you all in advance

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ImageJ mailing list: http://imagej.nih.gov/ij/list.html

[Adrián Villalba]

Dear Herbie,

First of all, thank you for helping. Yes, the data is not derived from my
research but the expected results - the information i would try to obtain
from that - is just to have a more specific idea of the point-location
along the y-axis (not data manipulation per se). Does it require an
authorization? yet could it also be considered as manipulation? If it does
i am sorry for the question,

Thank you very much

2017-06-19 16:01 GMT+02:00 Herbie <[hidden email]>:

> Good day Adrián,
>
> the question arises where these graphs come from. If they are not from
> your own research, what must be expected, the question arises, whether you
> are authorized to use or manipulate them.
>
> In general it is not a good idea to extract data from plots.
>
> Regards
>
> Herbie
>
> :::::::::::::::::::::::::::::::::::::::::::::
> Am 19.06.17 um 15:29 schrieb Adrián Villalba:
>
>> Dear all,
>>
>> I have a graph (line graph) and i want to know he exact at the y-axis of
>> all the points of the graph. I think of tracing manually a line over the
>> line-graph. Then trace another line among the basal line of the x-axis
>> (the
>> zero in the y-axis) and then calculate the distance (in any unit, it does
>> not matter) between each point of the graph line and its parallel to the
>> basal line. It would give me the proportionality respect a reference point
>> to know each value of the graph line.
>>
>> The fact is that when i try to do it on ImageJ, i can only do it manually
>> (which means a huge amount of time) do you know if there is any
>> choice/tool
>> to perform this calculation along the line-graph?
>>
>>
>> Thank you all in advance
>>
>
> --
> ImageJ mailing list: http://imagej.nih.gov/ij/list.html
>

--

   - Adrián Villalba Felipe.
   https://es.linkedin.com/in/adrianvillalba

--
ImageJ mailing list: http://imagej.nih.gov/ij/list.html

[Swayne, Theresa C.]

Dear Adrian,

I haven’t tried these particular methods, but here are 2 possibilities that came up in a quick search.

Luke Miller’s blog:
http://lukemiller.org/index.php/2011/09/digitizing-data-from-old-figures-with-imagej/

From the archives of this list:
http://imagej.1557.x6.nabble.com/Extract-Data-From-a-Plot-td5017731.html

Regarding Herbie’s comment, I agree that digitizing a plot doesn’t give you the most *precise* data. But I disagree that it’s somehow *unethical* to extract data from a published graph.
After all, when we look at a published graph we are allowed to mentally extract the data — we might even even pull out a ruler and pencil to get a better idea of the curve’s value at some point. Nothing seems wrong with that to me.

Of course, even better would be to ask the author for the source data. If the graph is in the scientific literature and not too old, many researchers would be willing to share an electronic version.

Hope this helps,
Theresa

On Jun 19, 2017, at 10:01 AM, Herbie <[hidden email]<mailto:[hidden email]>> wrote:

Good day Adrián,

the question arises where these graphs come from. If they are not from your own research, what must be expected, the question arises, whether you are authorized to use or manipulate them.

In general it is not a good idea to extract data from plots.

Regards

Herbie

:::::::::::::::::::::::::::::::::::::::::::::
Am 19.06.17 um 15:29 schrieb Adrián Villalba:
Dear all,
I have a graph (line graph) and i want to know he exact at the y-axis of
all the points of the graph. I think of tracing manually a line over the
line-graph. Then trace another line among the basal line of the x-axis (the
zero in the y-axis) and then calculate the distance (in any unit, it does
not matter) between each point of the graph line and its parallel to the
basal line. It would give me the proportionality respect a reference point
to know each value of the graph line.
The fact is that when i try to do it on ImageJ, i can only do it manually
(which means a huge amount of time) do you know if there is any choice/tool
to perform this calculation along the line-graph?
Thank you all in advance

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------------------------------------
Theresa Swayne, Ph.D.
Manager
Confocal and Specialized Microscopy Shared Resource<http://hiccc.columbia.edu/research/sharedresources/confocal>

Herbert Irving Comprehensive Cancer Center
Columbia University Medical Center
1130 St. Nicholas Ave., Room 222A
New York, NY 10032
Phone: [phone]
[hidden email]<mailto:[hidden email]>


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[Mark Krebs-2]

Hi Adrián,

I wonder if the Figure Calibration tool might be useful to you?

http://www.astro.physik.uni-goettingen.de/~hessman/ImageJ/Figure_Calibration/

Also, if you select the line by Image>Adust>Threshold, perhaps combined with skeletonizing, you should be able to output the xy coordinates through Analyze>Tools>Save XY Coordinates.  If you can make use of the calibration from Figure_Calibration you may obtain the values you seek.

Cheers,
Mark

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[Michael Schmid]

Hi Adrián,

ImageJ has a tool for extracting data from line graphs:
Analyze>Tools>Analyze Line Graph

https://imagej.nih.gov/ij/docs/guide/146-30.html#toc-Subsubsection-30.14.3

In my experience, it works best if you remove all axes etc. or put a
rectangle around the data, well inside of the axes; it also helps to
apply a threshold to the image (needed especially for plots that come
from jpg-compressed figures like often seen in PDFs).
Unfortunately, it does not accept the axis calibration with offset as
you can set with Image>Adjust>Coordinates; and it has a bug that
reverses the x axis in some cases (possibly related to y axis is
calibration with y ascending in upwards direction.)

[Concerning the point raised by Herbie - I consider it perfectly
legitimate to extract data from other's work for some purposes, e.g. as
a review of a manuscript for consistency checks, for a review article
where you plot several sets of data that you can't get directly from the
authors into one plot (with proper citations, etc. Of course, the
purpose of this tool is NOT to steal data to present them as the own work!]

Michael
________________________________________________________________
On 19/06/2017 15:29, Adrián Villalba wrote:

> Dear all,
>
> I have a graph (line graph) and i want to know he exact at the y-axis of
> all the points of the graph. I think of tracing manually a line over the
> line-graph. Then trace another line among the basal line of the x-axis (the
> zero in the y-axis) and then calculate the distance (in any unit, it does
> not matter) between each point of the graph line and its parallel to the
> basal line. It would give me the proportionality respect a reference point
> to know each value of the graph line.
>
> The fact is that when i try to do it on ImageJ, i can only do it manually
> (which means a huge amount of time) do you know if there is any choice/tool
> to perform this calculation along the line-graph?

--
ImageJ mailing list: http://imagej.nih.gov/ij/list.html

[Herbie]

Dear Theresa and Michael,

please read carefully!

I wrote that "the question arises, whether you are authorized to use or
manipulate them".

I didn't write that is it’s somehow *unethical* or legitimate.

I perfectly agree that there are reasons to do so, e.g. for personal use
or if correctly cited, but even in these cases, and as suggested by
Theresa, I would try to get in contact with the authors/persons who
generated the plots and ask them for the numerical data.

Regards

Herbie

::::::::::::::::::::::::::::::::::::::::::::
Am 19.06.17 um 16:49 schrieb Michael Schmid:

> Hi Adrián,
>
> ImageJ has a tool for extracting data from line graphs:
> Analyze>Tools>Analyze Line Graph
>
> https://imagej.nih.gov/ij/docs/guide/146-30.html#toc-Subsubsection-30.14.3
>
> In my experience, it works best if you remove all axes etc. or put a
> rectangle around the data, well inside of the axes; it also helps to
> apply a threshold to the image (needed especially for plots that come
> from jpg-compressed figures like often seen in PDFs).
> Unfortunately, it does not accept the axis calibration with offset as
> you can set with Image>Adjust>Coordinates; and it has a bug that
> reverses the x axis in some cases (possibly related to y axis is
> calibration with y ascending in upwards direction.)
>
> [Concerning the point raised by Herbie - I consider it perfectly
> legitimate to extract data from other's work for some purposes, e.g. as
> a review of a manuscript for consistency checks, for a review article
> where you plot several sets of data that you can't get directly from the
> authors into one plot (with proper citations, etc. Of course, the
> purpose of this tool is NOT to steal data to present them as the own work!]
>
> Michael
> ________________________________________________________________
> On 19/06/2017 15:29, Adrián Villalba wrote:
>> Dear all,
>>
>> I have a graph (line graph) and i want to know he exact at the y-axis of
>> all the points of the graph. I think of tracing manually a line over the
>> line-graph. Then trace another line among the basal line of the x-axis
>> (the
>> zero in the y-axis) and then calculate the distance (in any unit, it does
>> not matter) between each point of the graph line and its parallel to the
>> basal line. It would give me the proportionality respect a reference
>> point
>> to know each value of the graph line.
>>
>> The fact is that when i try to do it on ImageJ, i can only do it manually
>> (which means a huge amount of time) do you know if there is any
>> choice/tool
>> to perform this calculation along the line-graph?
>
> --
> ImageJ mailing list: http://imagej.nih.gov/ij/list.html
>

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[Cammer, Michael]

The data are presented to other scientists.  The data have passed one round of peer review and now available for anybody to review.  That's the point of publishing.  

And if you want to go a step further and publish your reanalysis of already published data even from a copyrighted source, if you acknowledge this, and especially if it is a review or meta-analysis, you are firmly protected under the law.


=*===========================================================*=
 Michael Cammer, DART Microscopy Laboratory, NYU Langone Medical Center
    Cell:  914-309-3270 (this is for calling, not texting)    Office: Skirball 2nd Floor main office, back right
      http://ocs.med.nyu.edu/microscopy & http://microscopynotes.com/



-----Original Message-----
From: Adrián Villalba [mailto:[hidden email]]
Sent: Monday, June 19, 2017 10:15 AM
To: [hidden email]
Subject: Re: Calculating distances between multiple points along a line

Dear Herbie,

First of all, thank you for helping. Yes, the data is not derived from my research but the expected results - the information i would try to obtain from that - is just to have a more specific idea of the point-location along the y-axis (not data manipulation per se). Does it require an authorization? yet could it also be considered as manipulation? If it does i am sorry for the question,

Thank you very much

2017-06-19 16:01 GMT+02:00 Herbie <[hidden email]>:

> Good day Adrián,
>
> the question arises where these graphs come from. If they are not from
> your own research, what must be expected, the question arises, whether
> you are authorized to use or manipulate them.
>
> In general it is not a good idea to extract data from plots.
>
> Regards
>
> Herbie
>
> :::::::::::::::::::::::::::::::::::::::::::::
> Am 19.06.17 um 15:29 schrieb Adrián Villalba:
>
>> Dear all,
>>
>> I have a graph (line graph) and i want to know he exact at the y-axis
>> of all the points of the graph. I think of tracing manually a line
>> over the line-graph. Then trace another line among the basal line of
>> the x-axis (the zero in the y-axis) and then calculate the distance
>> (in any unit, it does not matter) between each point of the graph
>> line and its parallel to the basal line. It would give me the
>> proportionality respect a reference point to know each value of the
>> graph line.
>>
>> The fact is that when i try to do it on ImageJ, i can only do it
>> manually (which means a huge amount of time) do you know if there is
>> any choice/tool to perform this calculation along the line-graph?
>>
>>
>> Thank you all in advance
>>
>
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--

   - Adrián Villalba Felipe.
   https://urldefense.proofpoint.com/v2/url?u=https-3A__es.linkedin.com_in_adrianvillalba&d=DQIFaQ&c=j5oPpO0eBH1iio48DtsedbOBGmuw5jHLjgvtN2r4ehE&r=oU_05LztNstAydlbm5L5GDu_vAdjXk3frDLx_CqKkuo&m=kxxvhVZZNpe7dsQDcJyup38G165ILgn4oAD4iSNNB0s&s=a9kVe5L-2z5UlXVrD8ePo1njxUPm0g5gtMjRvJWCwPw&e= 

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