Standardize[list]
shifts and rescales the elements of list to have zero mean and unit sample variance.
Standardize[list,f1]
shifts the elements in list by f1[list] and rescales them to have unit sample variance.
Standardize[list,f1,f2]
shifts by f1[list] and scales by f2[list].
Standardize
Standardize[list]
shifts and rescales the elements of list to have zero mean and unit sample variance.
Standardize[list,f1]
shifts the elements in list by f1[list] and rescales them to have unit sample variance.
Standardize[list,f1,f2]
shifts by f1[list] and scales by f2[list].
Details
- Standardize is also known as standard score or z-score.
- Standardize shifts by a location and rescales by a scale estimated from the elements of list.
- Standardize[list] is effectively (list-Mean[list])/StandardDeviation[list] for nonzero StandardDeviation[list].
- Standardize[list,f1] is effectively (list-f1[list])/StandardDeviation[list].
- Standardize[list,f1,f2] is effectively (list-f1[list])/f2[list].
- Common choices for f1 and f2 include:
-
Mean StandardDeviation zero mean and unit variance Mean 1& shift to mean 0 Median 1& shift by the median 0& StandardDeviation scale to have unit variance - Standardize handles both numerical and symbolic data.
- Standardize[{{x1,y1,…},{x2,y2,…},…}] effectively gives Transpose[{Standardize[{x1,x2,…}],Standardize[{y1,y2,…}],…}]. »
- Standardize works with SparseArray objects. »
Examples
open all close allBasic Examples (3)
Compute standard scores for data:
Standardize[{6.5, 3.8, 6.6, 5.7, 6.0, 6.4, 5.3}]{Mean[%], Variance[%]}Shift to have mean zero without scaling:
Standardize[{6.5, 3.8, 6.6, 5.7, 6.0, 6.4, 5.3}, Mean, 1&]{Mean[%], Variance[%]}Shift to Median with standard scaling:
Standardize[{6.5, 3.8, 6.6, 5.7, 6.0, 6.4, 5.3}, Median]{Mean[%], Median[%], Variance[%]}Shift to Median and scale by the InterquartileRange:
Standardize[{6.5, 3.8, 6.6, 5.7, 6.0, 6.4, 5.3}, Median, InterquartileRange]{Mean[%], Median[%], Variance[%], InterquartileRange[%]}Scope (6)
Standardize exact numeric data:
Standardize[Range[5]]Simplify[Standardize[{a, b, c}], {a, b, c}∈Reals]Standardize arbitrary‐precision data:
Standardize[N[Range[5], 20]]Standardize matrix data by column:
data = RandomReal[10, {10, 3}]Standardize[data]{Mean[%], Variance[%]}//ChopStandardize a TimeSeries:
ts = TemporalData[TimeSeries, {{{-0.8665776485273015, -0.6782110300443249, -0.5060797754514599,
-1.544046353019984, -1.548861927442106, 2.268550811187895, 0.7674881482532252,
-0.33377919918496324, -0.2623443622275032, -0.7898169819548497, -0.2 ... 5074135634818395, -7.993517405936538,
-8.11824028466883, -7.421451285939192, -7.617297878523265, -8.63887295423076}}, {{0, 78, 1}},
1, {"Continuous", 1}, {"Discrete", 1}, 1, {ValueDimensions -> 1, ResamplingMethod -> None}},
False, 10.1];{Mean[ts], Variance[ts]}res = Standardize[ts]{Mean[res], Variance[res]}//ChopStandardize a SparseArray:
sp = SparseArray[RandomInteger[1, 10]]Standardize[sp]Applications (3)
Shift normally distributed values to a standard normal:
data = RandomReal[NormalDistribution[5, 2], 500];sdata = Standardize[data];{ListPlot[Sort[data]], ListPlot[Sort[sdata]]}{Plot[InverseCDF[NormalDistribution[Mean[data], StandardDeviation[data]], q], {q, 0, 1}], Plot[InverseCDF[NormalDistribution[], q], {q, 0, 1}]}Center a data matrix by subtracting column means:
data = RandomReal[10, {5, 3}]cdata = Standardize[data, Mean, 1&]The mean of the centered data is 0:
Mean[cdata]//ChopThe covariance structure is unchanged:
MatrixForm /@ {Covariance[data], Covariance[cdata]}Use Standardize to centralize a TimeSeries:
ts = TemporalData[TimeSeries, {{{-0.8665776485273015, -0.6782110300443249, -0.5060797754514599,
-1.544046353019984, -1.548861927442106, 2.268550811187895, 0.7674881482532252,
-0.33377919918496324, -0.2623443622275032, -0.7898169819548497, -0.2 ... 5074135634818395, -7.993517405936538,
-8.11824028466883, -7.421451285939192, -7.617297878523265, -8.63887295423076}}, {{0, 78, 1}},
1, {"Continuous", 1}, {"Discrete", 1}, 1, {ValueDimensions -> 1, ResamplingMethod -> None}},
False, 10.1];Mean[ts]res = Standardize[ts, Mean, 1&]Mean[res]//ChopVariance[ts] == Variance[res]Properties & Relations (1)
Compute MeanDeviation using Standardize:
data = RandomReal[10, 20]Mean[Abs[Standardize[data, Mean, 1&]]]MeanDeviation[data]Similarly compute MedianDeviation:
Median[Abs[Standardize[data, Median, 1&]]]MedianDeviation[data]Related Guides
History
Text
Wolfram Research (2008), Standardize, Wolfram Language function, https://reference.wolfram.com/language/ref/Standardize.html.
CMS
Wolfram Language. 2008. "Standardize." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ref/Standardize.html.
APA
Wolfram Language. (2008). Standardize. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/Standardize.html
BibTeX
@misc{reference.wolfram_2026_standardize, author="Wolfram Research", title="{Standardize}", year="2008", howpublished="\url{https://reference.wolfram.com/language/ref/Standardize.html}", note=[Accessed: 13-June-2026]}
BibLaTeX
@online{reference.wolfram_2026_standardize, organization={Wolfram Research}, title={Standardize}, year={2008}, url={https://reference.wolfram.com/language/ref/Standardize.html}, note=[Accessed: 13-June-2026]}