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a=lambda b:lambda c:lambda d:lambda e:lambda f:0   # 48 bytes  (plain)
exec"a=`b:`c:`d:`e:`f:0".replace('`','lambda ')    # 47 bytes  (replace)
exec"a=%sb:%sc:%sd:%se:%sf:0"%(('lambda ',)*5)     # 46 bytes  (%)

a=lambda b:lambda c:lambda d:lambda e:lambda f:0   # 48 bytes  (plain)
exec"a=`b:`c:`d:`e:`f:0".replace('`','lambda ')    # 47 bytes  (replace)
exec"a=%sb:%sc:%sd:%se:%sf:0"%(('lambda ',)*5)     # 46 bytes  (%)

Use string substitution and eval to deal with long keyword like lambdas:

c="`x:`y:x"
exec c.replace('`', 'lambda ')

Can save strokes for a long run.

Alternatively, for a low number of things to replace, use string formatting:

compare:

exec"%sx:%sy:x"%(('lambda ',)*2)

and

exec"`x:`y:x".replace('`','lambda ')

String formatting takes two chars per thing to replace in, so replace wins if you need to replace a very large number of strings.

Problem, sequence point?

[ doThis(), doThat(1), getValue() ][2]

Use string substitution and exec to deal with long keywords like lambda that are repeated often in your code.

a=lambda b:lambda c:lambda d:lambda e:lambda f:0   # 48 bytes  (plain)
exec"a=`b:`c:`d:`e:`f:0".replace('`','lambda ')    # 47 bytes  (replace)
exec"a=%sb:%sc:%sd:%se:%sf:0"%(('lambda ',)*5)     # 46 bytes  (%)

The target string is very often 'lambda ', which is 7 bytes long. Suppose your code snippet contains n occurences of 'lambda ', and is s bytes long. Then:

• The plain option is s bytes long.
• The replace option is s - 6n + 29 bytes long.
• The % option is s - 5n + 22 + len(str(n)) bytes long.

From a plot of bytes saved over plain for these three options, we can see that:

• For n < 5 lambdas, you're better off not doing anything fancy at all.
• For n = 5, writing exec"..."%(('lambda ',)*5) saves 2 bytes, and is your best option.
• For n > 5, writing exec"...".replace('`','lambda ') is your best option.

For other cases, you can index the table below:

          1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 (occurences) 
       +---------------------------------------------------------
     3 |  -  -  -  -  -  -  -  -  -  -  -  -  -  -  r  r  r  r  r  
     4 |  -  -  -  -  -  -  -  -  -  r  r  r  r  r  r  r  r  r  r  
     5 |  -  -  -  -  -  -  -  r  r  r  r  r  r  r  r  r  r  r  r  
     6 |  -  -  -  -  -  r  r  r  r  r  r  r  r  r  r  r  r  r  r  
     7 |  -  -  -  -  %  r  r  r  r  r  r  r  r  r  r  r  r  r  r  
     8 |  -  -  -  %  %  r  r  r  r  r  r  r  r  r  r  r  r  r  r  
     9 |  -  -  -  %  %  r  r  r  r  r  r  r  r  r  r  r  r  r  r  
    10 |  -  -  %  %  %  r  r  r  r  r  r  r  r  r  r  r  r  r  r  
    11 |  -  -  %  %  %  r  r  r  r  r  r  r  r  r  r  r  r  r  r  
    12 |  -  -  %  %  %  r  r  r  r  r  r  r  r  r  r  r  r  r  r   r = replace
    13 |  -  -  %  %  %  r  r  r  r  r  r  r  r  r  r  r  r  r  r   % = string %
    14 |  -  %  %  %  %  r  r  r  r  r  r  r  r  r  r  r  r  r  r   - = do nothing
    15 |  -  %  %  %  %  r  r  r  r  r  r  r  r  r  r  r  r  r  r  
  (length)

For example, if the string lambda x,y: (length 11) occurs 3 times in your code, you're better off writing exec"..."%(('lambda x,y:',)*3).

Use string substitution and eval to deal with long keyword like lambdas:

c="`x:`y:x"
exec c.replace('`', 'lambda ')

Can save strokes for a long run

Problem, sequence point?

[ doThis(), doThat(1), getValue() ][2]

Use string substitution and eval to deal with long keyword like lambdas:

c="`x:`y:x"
exec c.replace('`', 'lambda ')

Can save strokes for a long run.

Alternatively, for a low number of things to replace, use string formatting:

compare:

exec"%sx:%sy:x"%(('lambda ',)*2)

and

exec"`x:`y:x".replace('`','lambda ')

String formatting takes two chars per thing to replace in, so replace wins if you need to replace a very large number of strings.

Problem, sequence point?

[ doThis(), doThat(1), getValue() ][2]