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Let's have a list:

d={0.00144444, 17.6365, 6.68574, 22.5808, -1.85266, -11.1622, 100.021, 1.}

I want to replace some parts of it by zeroes: Documentation of ReplacePart says:

ReplacePart[expr,{i,j,…}->new]

replaces the part at position {i,j,…}. but

ReplacePart[d, {1, 4, 5, 6, 7} -> 0.0]

produces

{0.00144444, 17.6365, 6.68574, 22.5808, -1.85266, -11.1622, 100.021, 1.}

and to make things stranger

ReplacePart[d, {{1}, 4, 5, 6, 7} -> 0.0]
{0., 17.6365, 6.68574, 0., 0., 0., 0., 1.}

but

ReplacePart[d, {1, {4}, 5, 6, 7} -> 0.0]

During evaluation of ReplacePart::pkspec1: The expression {4} cannot be used as a part specification.
(* {0.00144444, 17.6365, 6.68574, 22.5808, -1.85266, -11.1622, \
100.021, 1.} *)

Is this a bug? What is the proper syntax for ReplacePart in my situation?

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  • $\begingroup$ Anyway, I'm pretty sure that you want ReplacePart[data[[1, 1]], {{1}, {4}, {5}, {6}, {7}} -> 0.0] rather than ReplacePart[data[[1, 1]], {1, 4, 5, 6, 7} -> 0.0]? ReplacePart[data[[1, 1]], {{1}, 4, 5, 6, 7} -> 0.0] is weird, though. $\endgroup$ – march Aug 28 '17 at 17:57
  • $\begingroup$ @march I appologize, it was reckless, it has been corrected. The list to replace is 1 dimensional as was defined previously, the data[[1,1]] = d effectively. $\endgroup$ – leosenko Aug 28 '17 at 18:01
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    $\begingroup$ "replaces the part at position {i,j,…}" not "replaces parts at positions {i,j,…}" $\endgroup$ – Kuba♦ Aug 28 '17 at 18:05
  • $\begingroup$ @Kuba Is that supposed to mean at position [[i,j,...]]? {i,j,..} specifies a list, not a position, no? $\endgroup$ – leosenko Aug 28 '17 at 18:07
  • $\begingroup$ @leosenko I'd look for tips in Position docs because [[]] is from Part. $\endgroup$ – Kuba♦ Aug 28 '17 at 18:08
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Why not use Part itself?

d = {0.00144444, 17.6365, 6.68574, 22.5808, -1.85266, -11.1622, 100.021, 1.}; 
d[[{1, 4, 5, 6, 7}]] = 0;

Now:

d
{0, 17.6365, 6.68574, 0, 0, 0, 0, 1.}

If you really want to use ReplacePart:

ReplacePart[d, Thread[{1, 4, 5, 6, 7} -> {0, 0, 0, 0, 0}]]

gives the same answer.

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