assignment 1

assignment 1 iste python.txt

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+A1) Both the results are same, as list1+=[] is equivalent to list1 = list1 + [].
+      Result of print(lis1) is [�1�,�S�,�T�,�E]
+       But print(list2) is [�1�,�2�,�T�,�E�]
+In the end the identity of the list1 got changed due to addition of []. Hence changing the value of list2 doesn�t change the value of list1 anymore as both of them point to different objects now due to change in the id of both the lists.
+
+A2) for j in range(len(a)):
+...  swapped= False
+...  i=0
+...  while i<len(a)-1:
+...   if a[i]>a[i+1]:
+...    a[i],a[i+1] = a[i+1],a[i]
+...    swapped= True
+...   i=i+1
+...  if swapped== False:
+...   break
+
+list=9
+list1=[7,8,9,10]
+>>> for i in range(len(list1)):
+...  if list==list1[i]:
+...   print('element found at posn',i)
+...  else :
+...   i+=1
+...
+element found at posn 2
+
+A3) False, True
+Result of first print is False because == is a relational operator and it compares the ID of the 2 lists. It wont be true until both the lists point to the same object.
+       Result of second print is True because strings are not evaluated like arrays in python.]
+
+A4) 
+     i=0
+>>> list1=[7,8,9,10]
+>>> for i in range(len(list1)):
+...  if (list1[i]*list1[i]*list1[i])%3!=1:
+...   list1.remove(i)
+...  else:
+...   break