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I can't figure out how to add a constant value to a list. The list to which the data variable is bound was imported from a CSV file with 2 columns.

For example,

data = {{0.01, 0.02}, {0.08, 0.09}, {0.02, -0.05}, {0.03, 0.1}, ...}

How would I add a scalar value (e.g., +0.01) to the 2nd element of each pair?

The transformed data can be stored in a new variable data1.

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6 Answers 6

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you can try

#+{0,0.01}&/@{{0.01,0.02},{0.08,0.09},{0.02,-0.05},{0.03,0.1}}    

{{0.01, 0.03}, {0.08, 0.1}, {0.02, -0.04}, {0.03, 0.11}}

in other words

data1=#+{0,0.01}&/@data
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    $\begingroup$ Thank you. That's what I needed. $\endgroup$
    – Robin
    Apr 20, 2019 at 15:10
  • $\begingroup$ I'm glad I helped ! $\endgroup$
    – ZaMoC
    Apr 20, 2019 at 15:11
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Not the fastest, but simple to understand:

m = {{0.01,0.02},{0.08,0.09},{0.02,-0.05},{0.03,0.1}};
TranslationTransform[{0,.01}] @ m

{{0.01, 0.03}, {0.08, 0.1}, {0.02, -0.04}, {0.03, 0.11}}

Much faster is to use Transpose twice:

Transpose[Transpose[m] + {0., .01}]

{{0.01, 0.03}, {0.08, 0.1}, {0.02, -0.04}, {0.03, 0.11}}

Using Compile would be even faster.

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  • $\begingroup$ Excellent approaches. I will definitely try this out for very large sets of data, which I have. $\endgroup$
    – Robin
    Apr 20, 2019 at 15:13
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MapAt does exactly that:

data1 = MapAt[# + 0.01 &, data, {All, 2}]

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data = {{0.01, 0.02}, {0.08, 0.09}, {0.02, -0.05}, {0.03, 0.1}};

Using Threaded (new in 13.1)

data1 = data + Threaded[{0, 0.01}]

{{0.01, 0.03}, {0.08, 0.1}, {0.02, -0.04}, {0.03, 0.11}}

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Here is an alternative way to do it. I am not sure if it is fast though.

data = {{0.01, 0.02}, {0.08, 0.09}, {0.02, -0.05}, {0.03, 0.1}};
data[[All, 2]] += 0.01;
data

{{0.01, 0.03}, {0.08, 0.1}, {0.02, -0.04}, {0.03, 0.11}}

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data = {{0.01, 0.02}, {0.08, 0.09}, {0.02, -0.05}, {0.03, 0.1}};

Using SubsetMap:

SubsetMap[# + 0.01 &, data, {All, 2}]

(*{{0.01, 0.03}, {0.08, 0.1}, {0.02, -0.04}, {0.03, 0.11}}*)
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