2 3 Consider a list of values xs Find and return the value

2 3 # Consider a list of values, xs. Find and return the value at location index. 4 # If index is invalid for xs, return response instead. Remember that .get() 5 # for dictionaries works gracefully for non-existing keys. Here we are 6 # implementing a get() function for list type. 7 # -Parameters: 8 # -xs :: list of values of any length. 9 # -index :: an integer, specifying which value in the list to return. 10 # -response :: a python value, to be returned when index is invalid for the 11 # list. Defaults to None. 12 # -Return value: a value from xs at index or the pre-set response 13 # -suggestion : use try-except blocks in your solution. 14 # -Examples: 15# -get([\'a\',\'b\',\'c\'],0) 16 # -get((\'a\',\'b\',\'c\'],3) 17 # -get([\'a\",\'b\',\'c\'],4,\"oops\" ) --> \'oops \' 18 19 def get(xs, index, response None): 20 return \"not yet implemented\" 21 None

Solution

def get( xs, index, response=None):
   if( index >= len(xs) or index < 0):
       return response;
   return xs[index];

print get( [\'a\',\'b\',\'c\'], 0 );
print get( [\'a\',\'b\',\'c\'], 3 );
print get( [\'a\',\'b\',\'c\'], 4, \"oops\" );

def classify( input_string ):
   input_string= input_string.split(\" \");
   numbers= [];
   words = [];
   for x in input_string:
       try:
           x = int(x);
           numbers.append(x);
       except:
              if x <> \'\':
               words.append(x);
   return [ numbers, words ];

def shelve( inventory, product_list ):
   for p in product_list:
       try:
           inventory[ p[0] ] = inventory[ p[0] ] + p[1];
       except:
           inventory[ p[0] ] = p[1];
       if inventory[ p[0] ] < 0:
           raise ValueError(\'negative amount for \'+ p[0] );
   return None;

d = {\"kiwi\":999};
shelve(d,[(\"kiwi\",-2000)]);