Replace and ReplaceAll -- simple application
in [Mathematica]
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From: amannuc1 on 26 Dec 2009 19:09 I am doing some manipulations in Mathematica and end up with constructs such as "1." in the expressions. I want to convert these to "1" and thereby eliminate them. However, these examples don't seem to work. This is Mathematica 7 on OS X. In[40]:= Replace[1.(y1-1.yr),1.->1] Out[40]= 1. (y1-1. yr) (* Nothing happens *) In[39]:= ReplaceAll[1.(y1-1.yr),1.->1] Out[39]= y1-1. yr (* Only first one is replaced *) Any help is appreciated. Thanks. -Tony
From: Bob Hanlon on 27 Dec 2009 02:28 expr = 1. (y1 - 1. yr); expr /. x_?NumericQ -> Round[x] y1-yr expr // Rationalize y1-yr expr /. x_ :> Rationalize[x] y1-yr Bob Hanlon ---- amannuc1 <Anthony.J.Mannucci(a)jpl.nasa.gov> wrote: ============= I am doing some manipulations in Mathematica and end up with constructs such as "1." in the expressions. I want to convert these to "1" and thereby eliminate them. However, these examples don't seem to work. This is Mathematica 7 on OS X. In[40]:= Replace[1.(y1-1.yr),1.->1] Out[40]= 1. (y1-1. yr) (* Nothing happens *) In[39]:= ReplaceAll[1.(y1-1.yr),1.->1] Out[39]= y1-1. yr (* Only first one is replaced *) Any help is appreciated. Thanks. -Tony
From: Scot T. Martin on 27 Dec 2009 02:30 The problem is that "-1." is not "1". Also, you need to use the level specification. In: Replace[1. (y1 - 1. yr), {1. -> 1, -1. -> -1}, Infinity] Out: y1-yr On Sat, 26 Dec 2009, amannuc1 wrote: > I am doing some manipulations in Mathematica and end up with > constructs such as "1." in the expressions. I want to convert these to > "1" and thereby eliminate them. However, these examples don't seem to > work. This is Mathematica 7 on OS X. > > In[40]:= > Replace[1.(y1-1.yr),1.->1] > Out[40]= 1. (y1-1. yr) (* Nothing happens *) > > In[39]:= > ReplaceAll[1.(y1-1.yr),1.->1] > Out[39]= y1-1. yr (* Only first one is replaced *) > > Any help is appreciated. Thanks. > > -Tony > >
From: David Park on 27 Dec 2009 02:30 expr = 1. (y1 - 1. yr); Look at the FullForm. expr // FullForm Times[1.`,Plus[y1,Times[-1.`,yr]]] There is a 1. And a -1. and you must have a replacement for each of them. expr /. {1. -> 1, -1. -> -1} y1 - yr expr /. a : (1. | -1.) -> Sign[a] y1 - yr Another possible construction is: expr // Rationalize y1 - yr I don't know how you arrived at the initial expression, but if possible you could try to enter only exact expressions (avoiding numbers with decimal points, called approximate numbers). Enter approximate numbers only when you are absolutely forced to. Try to keep all equations symbolic, free of approximate numbers. Also keep them free of units. Then once you have solved the equations (if you can do that) then substitute data values with approximate numbers and units. David Park djmpark(a)comcast.net http://home.comcast.net/~djmpark/ From: amannuc1 [mailto:Anthony.J.Mannucci(a)jpl.nasa.gov] I am doing some manipulations in Mathematica and end up with constructs such as "1." in the expressions. I want to convert these to "1" and thereby eliminate them. However, these examples don't seem to work. This is Mathematica 7 on OS X. In[40]:= Replace[1.(y1-1.yr),1.->1] Out[40]= 1. (y1-1. yr) (* Nothing happens *) In[39]:= ReplaceAll[1.(y1-1.yr),1.->1] Out[39]= y1-1. yr (* Only first one is replaced *) Any help is appreciated. Thanks. -Tony
From: DrMajorBob on 27 Dec 2009 02:30
1. (y1 - 1. yr) // Rationalize y1 - yr Bobby On Sat, 26 Dec 2009 18:06:43 -0600, amannuc1 <Anthony.J.Mannucci(a)jpl.nasa.gov> wrote: > I am doing some manipulations in Mathematica and end up with > constructs such as "1." in the expressions. I want to convert these to > "1" and thereby eliminate them. However, these examples don't seem to > work. This is Mathematica 7 on OS X. > > In[40]:= > Replace[1.(y1-1.yr),1.->1] > Out[40]= 1. (y1-1. yr) (* Nothing happens *) > > In[39]:= > ReplaceAll[1.(y1-1.yr),1.->1] > Out[39]= y1-1. yr (* Only first one is replaced *) > > Any help is appreciated. Thanks. > > -Tony > -- DrMajorBob(a)yahoo.com |