There are some basic instructions about Composition on this site, but I'm still confused on how to use it when the functions I am composing have two parameter, such as:

t := RandomImage[1, {50, 50}, ColorSpace -> "RGB"]
picList = Array[t &, 3];
ImageResize[#, 500] &@*ColorConvert[#, "Grayscale"] & /@ picList

Mathematica graphics

But when the composition functions just have one parameter, the behavior is more predictable:

hh@*ff /@ picList

Mathematica graphics

Can anybody explain this strange behavior?


1 Answer 1


& has a very low operator precedence, in particular it's lower than @*'s precedence. That means that the right-hand & is actually defining an unnamed function with the body:

ImageResize[#, 500] & @* ColorConvert[#, "Grayscale"]

You can fix this by wrapping the right-hand function in parentheses:

ImageResize[#, 500] & @* (ColorConvert[#, "Grayscale"] &) /@ piclist

Or both of them for consistency:

(ImageResize[#, 500] &) @* (ColorConvert[#, "Grayscale"] &) /@ piclist

Of course, as you mentioned in a comment, that doesn't help much compared with

(ImageResize[#, 500] &) /@ (ColorConvert[#, "Grayscale"] &) /@ piclist

If you want to avoid the parentheses altogether, you could still define a pure function via the Function, but of course, that's even less terse, and technically you've just replaced parentheses with square brackets:

Function[ImageResize[#, 500]] @* Function[ColorConvert[#, "Grayscale"]] /@ piclist

You can also use the full form of @* instead:

Composition[ImageResize[#, 500] &, ColorConvert[#, "Grayscale"] &] /@ piclist

Alternatively, you could apply both transformations in a single function:

ImageResize[ColorConvert[#, "Grayscale"], 500] & /@ piclist

Or use infix notation for a nicer reading order:

# ~ColorConvert~ "Grayscale" ~ImageResize~ 500 & /@ piclist
  • $\begingroup$ Applaud for the last form :) $\endgroup$
    – yode
    Mar 29, 2016 at 12:17
  • $\begingroup$ Or we can use Composition[ImageResize[#,500]&,ColorConvert[#,"Grayscale"]&]/@picList to aviod the bracket $\endgroup$
    – yode
    Mar 29, 2016 at 12:25
  • $\begingroup$ @yode Oh, good point, I'll add that for completeness. $\endgroup$ Mar 29, 2016 at 12:33
  • 2
    $\begingroup$ For something this elaborate, I'd go with the penultimate snippet. Not worth having to try to look up precedence just to have compact code. $\endgroup$ Mar 29, 2016 at 13:04

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