Answer:
1. DBMS
C. <em>A storage system that provides efficient access to randomly chosen items</em>
G. <em>Performs database operations requested by application software</em>
2. data mining
B. <em>The process of extracting hidden information</em>
3. hash file
A. <em>A. means of locating a particular record within a file</em>
4. index key field
F. <em>An item used to identify records uniquely</em>
5. locking protocol
E. <em>A system to guard against database errors due to performing transactions concurrently</em>
6. relation
D. <em>A structural unit (with rows and columns) in a popular database model</em>
7. roll back schema
I. <em>A "road map" of a particular database's design</em>
J. <em>To "unwind" a transaction</em>
8. SQL
H. <em>A popular language that implements relational database operations.</em>
Answer:
Most likely C
Explanation:
Blueprint is the plan so she makes it 1st.
Building the roof is after the deck.
The deck is after the blueprint.
After it's all done she looks at it and thinks about what went wrong at then end.
If there are no mistakes it's B.
Hope this answers your question :).
Answer:
from collections import Counter
def anagram(dictionary, query):
newList =[]
for element in dictionary:
for item in query:
word = 0
count = 0
for i in [x for x in item]:
if i in element:
count += 1
if count == len(item):
newList.append(item)
ans = list()
for point in Counter(newList).items():
ans.append(point)
print(ans)
mylist = ['jack', 'run', 'contain', 'reserve','hack','mack', 'cantoneese', 'nurse']
setter = ['ack', 'nur', 'can', 'con', 'reeve', 'serve']
anagram(mylist, setter)
Explanation:
The Counter class is used to create a dictionary that counts the number of anagrams in the created list 'newList' and then the counter is looped through to append the items (tuple of key and value pairs) to the 'ans' list which is printed as output.
Code:
def myAppend( str, ch ):
# Return a new string that is like str but with
# character ch added at the end
return str + ch
def myCount( str, ch ):
# Return the number of times character ch appears
# in str.
# initiaalizing count with 0
count = 0
# iterating over every characters present in str
for character in str:
# incrementing count by 1 if character == ch
if character == ch:
count += 1
# returning count
return count
def myExtend( str1, str2 ):
# Return a new string that contains the elements of
# str1 followed by the elements of str2, in the same
# order they appear in str2.
# concatenating both strings and returning its result
return str1 + str2
def myMin( str ):
# Return the character in str with the lowest ASCII code.
# If str is empty, print "Empty string: no min value"
# and return None.
if str == "":
print("Empty string: no min value")
return None
# storing first character from str in char
char = str[0]
# iterating over every characters present in str
for character in str:
# if current character is lower than char then
# assigning char with current character
if character < char:
char = character
# returning char
return char
def myInsert( str, i, ch ):
# Return a new string like str except that ch has been
# inserted at the ith position. I.e., the string is now
# one character longer than before.
# Print "Invalid index" if
# i is greater than the length of str and return None.
if i > len(str):
print("Invalid index")
return None
# str[:i] gives substring starting from 0 and upto ith position
# str[i:] gives substring starting from i and till last position
# returning the concatenated result of all three
return str[:i]+ch+str[i:]
def myPop( str, i ):
# Return two results:
# 1. a new string that is like str but with the ith
# element removed;
# 2. the value that was removed.
# Print "Invalid index" if i is greater than or
# equal to len(str), and return str unchanged and None
if i >= len(str):
print("Invalid index")
return str, None
# finding new string without ith character
new_str = str[:i] + str[i+1:]
# returning new_str and popped character
return new_str, str[i]
def myFind( str, ch ):
# Return the index of the first (leftmost) occurrence of
# ch in str, if any. Return -1 if ch does not occur in str.
# finding length of the string
length = len(str)
# iterating over every characters present in str
for i in range(length):
# returning position i at which character was found
if str[i]==ch:
return i
# returning -1 otherwise
return -1
def myRFind( str, ch ):
# Return the index of the last (rightmost) occurrence of
# ch in str, if any. Return -1 if ch does not occur in str.
# finding length of the string
length = len(str)
# iterating over every characters present in str from right side
for i in range(length-1, 0, -1):
# returning position i at which character was found
if str[i]==ch:
return i
# returning -1 otherwise
return -1
def myRemove( str, ch ):
# Return a new string with the first occurrence of ch
# removed. If there is none, return str.
# returning str if ch is not present in str
if ch not in str:
return str
# finding position of first occurence of ch in str
pos = 0
for char in str:
# stopping loop if both character matches
if char == ch:
break
# incrementing pos by 1
pos += 1
# returning strig excluding first occurence of ch
return str[:pos] + str[pos+1:]
def myRemoveAll( str, ch ):
# Return a new string with all occurrences of ch.
# removed. If there are none, return str.
# creating an empty string
string = ""
# iterating over each and every character of str
for char in str:
# if char is not matching with ch then adding it to string
if char!=ch:
string += char
# returning string
return string
def myReverse( str ):
# Return a new string like str but with the characters
# in the reverse order.
return str[::-1]
