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Using DeleteCases with a defined function with two arguments as a pattern
Using DeleteCases with multiple AstronomicalData propertiesRemoving some data and replacing them with new onesAdvanced SQL-type select statements (filtering) on DatasetsFiltering outliersHow to remove outliers from dataPattern-matching Function with conditions on arbitrary number of argumentsHow to delete matrix elements where the conditions apply to more than one dimensional elementChange the point size in ListPlotHow do I remove particular $x$ and $y$ values from a list of data points?Cleaning away data points which are enveloped within a function
$begingroup$
I am trying to manipulate some large datasets in Mathematica. I have plotted the data, and I would like to be able to easily remove points from the dataset, and it's fairly easy to identify the x-values of the points I need to remove. Let's say I need to remove the data point at an x-value of 1.1. To do this, I could do something like this:
BadData[entry_] := MatchQ[entry, 1.1, _?NumberQ]
goodData = DeleteCases[data, _?BadData]
However, I have 6 other data sets, all of which I have to remove different outliers (not at 1.1). I'd like to define a function such as
BadData[entry_, bad_] := MatchQ[entry, bad, _?NumberQ]
because then for bad I can just put whatever the x-value is of the point I need to remove. I tested the function out by figuring out that the data point with 1.1 as the x-value is the 7th in data, and running
BadData[data[[7]], 1.1]
returns True. So I know the function works the way I want it to, even though I know this might not be the "prettiest" way to do this.
What I can't figure out is how to use this version of baddata as a pattern in DeleteCases because I don't know how to specify the second variable, which would be 1.1 for this set. Is this possible with the setup I have currently?
list-manipulation pattern-matching filtering dataset argument-patterns
asked 1 hour ago
nwellsnwells
82
New contributor
nwells is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am trying to manipulate some large datasets in Mathematica. I have plotted the data, and I would like to be able to easily remove points from the dataset, and it's fairly easy to identify the x-values of the points I need to remove. Let's say I need to remove the data point at an x-value of 1.1. To do this, I could do something like this:
BadData[entry_] := MatchQ[entry, 1.1, _?NumberQ]
goodData = DeleteCases[data, _?BadData]
However, I have 6 other data sets, all of which I have to remove different outliers (not at 1.1). I'd like to define a function such as
BadData[entry_, bad_] := MatchQ[entry, bad, _?NumberQ]
because then for bad I can just put whatever the x-value is of the point I need to remove. I tested the function out by figuring out that the data point with 1.1 as the x-value is the 7th in data, and running
BadData[data[[7]], 1.1]
returns True. So I know the function works the way I want it to, even though I know this might not be the "prettiest" way to do this.
What I can't figure out is how to use this version of baddata as a pattern in DeleteCases because I don't know how to specify the second variable, which would be 1.1 for this set. Is this possible with the setup I have currently?
list-manipulation pattern-matching filtering dataset argument-patterns
asked 1 hour ago
nwellsnwells
82
New contributor
nwells is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Try this:DeleteCases[data, _?(BadData[#, 1.1] &)]
$endgroup$
– Anjan Kumar
1 hour ago
add a comment |
$begingroup$
I am trying to manipulate some large datasets in Mathematica. I have plotted the data, and I would like to be able to easily remove points from the dataset, and it's fairly easy to identify the x-values of the points I need to remove. Let's say I need to remove the data point at an x-value of 1.1. To do this, I could do something like this:
BadData[entry_] := MatchQ[entry, 1.1, _?NumberQ]
goodData = DeleteCases[data, _?BadData]
However, I have 6 other data sets, all of which I have to remove different outliers (not at 1.1). I'd like to define a function such as
BadData[entry_, bad_] := MatchQ[entry, bad, _?NumberQ]
because then for bad I can just put whatever the x-value is of the point I need to remove. I tested the function out by figuring out that the data point with 1.1 as the x-value is the 7th in data, and running
BadData[data[[7]], 1.1]
returns True. So I know the function works the way I want it to, even though I know this might not be the "prettiest" way to do this.
What I can't figure out is how to use this version of baddata as a pattern in DeleteCases because I don't know how to specify the second variable, which would be 1.1 for this set. Is this possible with the setup I have currently?
list-manipulation pattern-matching filtering dataset argument-patterns
asked 1 hour ago
nwellsnwells
82
New contributor
nwells is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am trying to manipulate some large datasets in Mathematica. I have plotted the data, and I would like to be able to easily remove points from the dataset, and it's fairly easy to identify the x-values of the points I need to remove. Let's say I need to remove the data point at an x-value of 1.1. To do this, I could do something like this:
BadData[entry_] := MatchQ[entry, 1.1, _?NumberQ]
goodData = DeleteCases[data, _?BadData]
However, I have 6 other data sets, all of which I have to remove different outliers (not at 1.1). I'd like to define a function such as
BadData[entry_, bad_] := MatchQ[entry, bad, _?NumberQ]
because then for bad I can just put whatever the x-value is of the point I need to remove. I tested the function out by figuring out that the data point with 1.1 as the x-value is the 7th in data, and running
BadData[data[[7]], 1.1]
returns True. So I know the function works the way I want it to, even though I know this might not be the "prettiest" way to do this.
What I can't figure out is how to use this version of baddata as a pattern in DeleteCases because I don't know how to specify the second variable, which would be 1.1 for this set. Is this possible with the setup I have currently?
list-manipulation pattern-matching filtering dataset argument-patterns
list-manipulation pattern-matching filtering dataset argument-patterns
asked 1 hour ago
nwellsnwells
82
New contributor
nwells is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
nwellsnwells
82
New contributor
nwells is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
nwellsnwells
82
New contributor
nwells is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
nwellsnwells
82
asked 1 hour ago
nwellsnwells
82
82
New contributor
nwells is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
nwells is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
nwells is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Try this:DeleteCases[data, _?(BadData[#, 1.1] &)]
$endgroup$
– Anjan Kumar
1 hour ago
add a comment |
1
$begingroup$
Try this:DeleteCases[data, _?(BadData[#, 1.1] &)]
$endgroup$
– Anjan Kumar
1 hour ago
1
1
$begingroup$
Try this:
DeleteCases[data, _?(BadData[#, 1.1] &)]$endgroup$
– Anjan Kumar
1 hour ago
$begingroup$
Try this:
DeleteCases[data, _?(BadData[#, 1.1] &)]$endgroup$
– Anjan Kumar
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use an implicit function more explicitly, as
data = 1, 5, 1.1, 6, 2, 7;
BadData[entry_, bad_] := MatchQ[entry, bad, _?NumberQ]
DeleteCases[data, _?(BadData[#, 1.1] &)]
1, 5, 2, 7
answered 1 hour ago
KagaratschKagaratsch
4,94731350
$endgroup$
add a comment |
$begingroup$
You can also define BadData as a pure function:
BadData[bad_] := First[#] == bad&
Then:
DeleteCases[data, _?(BadData[1.1])]
1, 5, 2, 7
Another possibility is:
BadData[bad_] := EqualTo[bad] @* First
answered 1 hour ago
Carl WollCarl Woll
76.6k3100201
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use an implicit function more explicitly, as
data = 1, 5, 1.1, 6, 2, 7;
BadData[entry_, bad_] := MatchQ[entry, bad, _?NumberQ]
DeleteCases[data, _?(BadData[#, 1.1] &)]
1, 5, 2, 7
answered 1 hour ago
KagaratschKagaratsch
4,94731350
$endgroup$
add a comment |
$begingroup$
You can use an implicit function more explicitly, as
data = 1, 5, 1.1, 6, 2, 7;
BadData[entry_, bad_] := MatchQ[entry, bad, _?NumberQ]
DeleteCases[data, _?(BadData[#, 1.1] &)]
1, 5, 2, 7
answered 1 hour ago
KagaratschKagaratsch
4,94731350
$endgroup$
add a comment |
$begingroup$
You can use an implicit function more explicitly, as
data = 1, 5, 1.1, 6, 2, 7;
BadData[entry_, bad_] := MatchQ[entry, bad, _?NumberQ]
DeleteCases[data, _?(BadData[#, 1.1] &)]
1, 5, 2, 7
answered 1 hour ago
KagaratschKagaratsch
4,94731350
$endgroup$
You can use an implicit function more explicitly, as
data = 1, 5, 1.1, 6, 2, 7;
BadData[entry_, bad_] := MatchQ[entry, bad, _?NumberQ]
DeleteCases[data, _?(BadData[#, 1.1] &)]
1, 5, 2, 7
answered 1 hour ago
KagaratschKagaratsch
4,94731350
answered 1 hour ago
KagaratschKagaratsch
4,94731350
answered 1 hour ago
KagaratschKagaratsch
4,94731350
answered 1 hour ago
KagaratschKagaratsch
4,94731350
4,94731350
add a comment |
add a comment |
$begingroup$
You can also define BadData as a pure function:
BadData[bad_] := First[#] == bad&
Then:
DeleteCases[data, _?(BadData[1.1])]
1, 5, 2, 7
Another possibility is:
BadData[bad_] := EqualTo[bad] @* First
answered 1 hour ago
Carl WollCarl Woll
76.6k3100201
$endgroup$
add a comment |
$begingroup$
You can also define BadData as a pure function:
BadData[bad_] := First[#] == bad&
Then:
DeleteCases[data, _?(BadData[1.1])]
1, 5, 2, 7
Another possibility is:
BadData[bad_] := EqualTo[bad] @* First
answered 1 hour ago
Carl WollCarl Woll
76.6k3100201
$endgroup$
add a comment |
$begingroup$
You can also define BadData as a pure function:
BadData[bad_] := First[#] == bad&
Then:
DeleteCases[data, _?(BadData[1.1])]
1, 5, 2, 7
Another possibility is:
BadData[bad_] := EqualTo[bad] @* First
answered 1 hour ago
Carl WollCarl Woll
76.6k3100201
$endgroup$
You can also define BadData as a pure function:
BadData[bad_] := First[#] == bad&
Then:
DeleteCases[data, _?(BadData[1.1])]
1, 5, 2, 7
Another possibility is:
BadData[bad_] := EqualTo[bad] @* First
answered 1 hour ago
Carl WollCarl Woll
76.6k3100201
answered 1 hour ago
Carl WollCarl Woll
76.6k3100201
answered 1 hour ago
Carl WollCarl Woll
76.6k3100201
answered 1 hour ago
Carl WollCarl Woll
76.6k3100201
76.6k3100201
add a comment |
add a comment |
nwells is a new contributor. Be nice, and check out our Code of Conduct.
nwells is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Try this:
DeleteCases[data, _?(BadData[#, 1.1] &)]$endgroup$
– Anjan Kumar
1 hour ago