Represents a histogram (map from values to integer frequencies).
Function or value | Description |
Full Usage:
add equalBandwidthOrNominal histA histB
Parameters:
bool
-
Is the binwidth equal for both frequencies? For nominal data set to true.
histA : Map<'a, ^value>
-
Frequency map A
histB : Map<'a, ^value>
-
Frequency map B
Returns: Map<'a, ^value>
New frequency map that results from merged maps histA and histB. Values from keys that are present in both maps are handled by f
Modifiers: inline |
Merges two histograms into a single histogram. If a key exists in both histograms, the value from histA is added to the value of histB. When applied to continuous data the bandwidths must be equal!
|
|
Example
|
Full Usage:
create bandwidth data
Parameters:
float
-
data : seq<float>
-
Returns: Map<float, int>
|
Example
|
|
Given the list [a,b,a,c,b,b], produce a map {a:2, b:3, c:1} which contains the count of each unique item in the list
Example
|
|
Example
|
|
Example
|
|
Example
|
|
Example
|
|
Example
|
Full Usage:
merge equalBandwidthOrNominal histA histB
Parameters:
bool
-
Is the binwidth equal for both frequencies? For nominal data set to true.
histA : Map<'a, 'value>
-
Frequency map A
histB : Map<'a, 'value>
-
Frequency map B
Returns: Map<'a, 'value>
New frequency map that results from merged maps histA and histB.
|
Merges two histograms into a single histogram. If a key exists in both histograms, the value in histA is superseded by the value in histB. When applied to continuous data the bandwidths must be equal!This function is not commutative! (merge a b) is not equal to (merge b a)
|
Full Usage:
mergeBy equalBandwidthOrNominal f histA histB
Parameters:
bool
-
Is the binwidth equal for both frequencies? For nominal data set to true.
f : 'value -> 'value -> 'value
-
Function to transform values if key is present in both histograms. `mapA-value → mapB-value → newValue`
histA : Map<'a, 'value>
histB : Map<'a, 'value>
Returns: Map<'a, 'value>
New frequency map that results from merged maps mapA and mapB. Values from keys that are present in both maps are handled by f
|
Merges two histograms into a single histogram. If a key exists in both maps, the value is determined by f with the first value being from mapA and the second originating from mapB. When applied to continuous data the bandwidths must be equal!This function is not commutative! (mergeBy f a b) is not equal to (mergeBy f b a)
|
Full Usage:
subtract equalBandwidthOrNominal histA histB
Parameters:
bool
-
Is the binwidth equal for both frequencies? For nominal data set to true.
histA : Map<'a, ^value>
-
Frequency map A
histB : Map<'a, ^value>
-
Frequency map B
Returns: Map<'a, ^value>
Modifiers: inline |
Merges two histograms into a single histogram. If a key exists in both histograms, the value from histB is subtracted from the value of histA. When applied to continuous data the bandwidths must be equal!This function is not commutative! (subtract a b) is not equal to (subtract b a)
|
Example
|