Each week I import a dataset, and every week the dataset differs in length (the number of records). After the import I need to do some transformations. One of these transformations is the ReplaceAll.

problem: when the number of records is the same length of the number of parameters in de transformations, the ReplaceAll won't give the desired output.

For example:

data1 = {{"A1", "B1", "C1", "D1", "E1"}, {"A2", "B2", "C2", "D2", 
   "E2"}, {"A3", "B3", "C3", "D3", "E3"}, {"A4", "B4", "C4", "D4", 
   "E4"}, {"A5", "B5", "C5", "D5", "E5"}};

data1 /. {a_ , b_, c_ , d_ , e_} -> {"T", a , b, c, d, e}

The output is:

{"T", {"A1", "B1", "C1", "D1", "E1"}, {"A2", "B2", "C2", "D2", "E2"}, {"A3", "B3", "C3", "D3", "E3"}, {"A4", "B4", "C4", "D4", "E4"}, {"A5", "B5", "C5", "D5", "E5"}}

When the number of records differ from the number of parameters in the ReplaceAll function, like this:

data2 = {{"A1", "B1", "C1", "D1", "E1"}, {"A2", "B2", "C2", "D2", 
   "E2"}, {"A3", "B3", "C3", "D3", "E3"}, {"A4", "B4", "C4", "D4", 

data2 /. {a_ , b_, c_ , d_ , e_} -> {"T", a , b, c, d, e}

Now I get the output I want.

{{"T", "A1", "B1", "C1", "D1", "E1"}, {"T", "A2", "B2", "C2", "D2", "E2"}, {"T", "A3", "B3", "C3", "D3", "E3"}, {"T", "A4", "B4", "C4", "D4", "E4"}}

The same is truth for a list with six records:

data3 = {{"A1", "B1", "C1", "D1", "E1"}, {"A2", "B2", "C2", "D2", 
   "E2"}, {"A3", "B3", "C3", "D3", "E3"}, {"A4", "B4", "C4", "D4", 
   "E4"}, {"A5", "B5", "C5", "D5", "E5"}, {"A6", "B6", "C6", "D6", 

data3 /. {a_ , b_, c_ , d_ , e_} -> {"T", a , b, c, d, e}

{{"T", "A1", "B1", "C1", "D1", "E1"}, {"T", "A2", "B2", "C2", "D2", "E2"}, {"T", "A3", "B3", "C3", "D3", "E3"}, {"T", "A4", "B4", "C4", "D4", "E4"}, {"T", "A5", "B5", "C5", "D5", "E5"}, {"T", "A6", "B6", "C6", "D6", "E6"}}

Who has a suggestion for this problem?


3 Answers 3


Use Replace with a level specification instead:

data1 = {


{{"T", "A1", "B1", "C1", "D1", "E1"}, {"T", "A2", "B2", "C2", "D2", "E2"}, {"T", "A3", "B3", "C3", "D3", "E3"}, {"T", "A4", "B4", "C4", "D4", "E4"}, {"T", "A5", "B5", "C5", "D5", "E5"}}


Try Prepend

Prepend[#, "T"] & /@ data1

{{"T", "A1", "B1", "C1", "D1", "E1"}, {"T", "A2", "B2", "C2", "D2", "E2"}, {"T", "A3", "B3", "C3", "D3", "E3"}, {"T", "A4", "B4", "C4", "D4", "E4"}, {"T", "A5", "B5", "C5", "D5", "E5"}}

Besides, {a_ , b_, c_ , d_ , e_} -> {"T", a , b, c, d, e} can be written as {a__} -> {"T", a} , if the list is not empty.

Or you can try

# /. {a__} :> {"T", a} & /@ {{"A1", "B1", "C1", "D1", "E1"}, {"A2", 
            "B2", "C2", "D2", "E2"}, {"A3", "B3", "C3", "D3", "E3"}, {"A4", 
            "B4", "C4", "D4", "E4"}, {"A5", "B5", "C5", "D5", "E5"}}

It is because _ operator starts at top where it found List and only if number of elements in list are same as _(as a requirement of _), it will recognize it otherwise it will go down to elements of each list. Say you do a small test,

data2 /. {a_, b_, c_, d_, e_} /; ListQ[a] -> {"T", a, b, c, d, e}

It returns,

{{"A1", "B1", "C1", "D1", "E1"}, {"A2", "B2", "C2", "D2", 
  "E2"}, {"A3", "B3", "C3", "D3", "E3"}, {"A4", "B4", "C4", "D4", 

Nothing processed because pattern didn't match.

To solve it, something on these line,

data2 //. {a___, {b_, c_, d_, e_, f_}, 
   g___} :> {a, {"T", b, c, d, e, f}, g}

{{"T", "A1", "B1", "C1", "D1", "E1"}, {"T", "A2", "B2", "C2", "D2", 
  "E2"}, {"T", "A3", "B3", "C3", "D3", "E3"}, {"T", "A4", "B4", "C4", 
  "D4", "E4"}}
  • $\begingroup$ Oke, thanks, But do you have a solution for my problem $\endgroup$ Nov 9, 2020 at 13:28

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