reduce expression to test sublist

Jussi Piitulainen jpiitula at
Sat Jan 5 21:41:24 CET 2013

Asim writes:

> Hi All
> The following reduce expression checks if every element of list lst1
> is present in list lst2.  It works as expected for integer lists but
> for lists of strings, it always returns False.
>    reduce( lambda x,y: (x in lst2) and (y in lst2), lst1)

Possibly this:

  >>> True in [3, 1, 4]
  >>> True in ["3", "1", "4"]

Since reduce(f, [a, b, c]) == f(f(a,b),c), your x will be True or
False except in the innermost call where it is the first element of
the list being reduced.

It doesn't really work with integers either. Only in certain special
cases: very short lst1, or True in lst2.

> Moreover, for the lists of strings the following for-loop gives
> correct results when the above reduce expression doesn't.
>    isSublist = True
>    for i in lst1:
>       isSublist = isSublist and (i in lst2)
>       if not isSublist:
>          isSublist = False
>          break
> Can someone help me understand why?

Consider reduce(lambda x, y: x and (y in whatever), ys, True).

Maybe also consider all(y in whatever for y in ys).

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