python - String vs list membership check -

so i'm wondering why this:

'alpha' in 'alphanumeric' 

is true,

list('alpha') in list('alphanumeric') 

is false.

why x in s succeed when x substring of s, x in l doesn't when x sublist of l?

when use list function iterable, new list object created elements iterable individual elements in list.

in case, strings valid python iterables, so

>>> list('alpha') ['a', 'l', 'p', 'h', 'a'] >>> list('alphanumeric') ['a', 'l', 'p', 'h', 'a', 'n', 'u', 'm', 'e', 'r', 'i', 'c'] 

so, checking if 1 list sublist of list.

in python strings have in operator check if 1 string part of string. other collections, can use individual members. quoting documentation,

the operators in , not in test collection membership. x in s evaluates true if x member of collection s, , false otherwise. x not in s returns negation of x in s. collection membership test has traditionally been bound sequences; object member of collection if collection sequence , contains element equal object. however, make sense many other object types support membership tests without being sequence. in particular, dictionaries (for keys) , sets support membership testing.

for list , tuple types, x in y true if , if there exists index i such x == y[i] true.

for unicode , string types, x in y true if , if x substring of y. equivalent test y.find(x) != -1. note, x , y need not same type; consequently, u'ab' in 'abc' return true. empty strings considered substring of other string, "" in "abc" return true.