Implement the function unique that takes as a parameter a two-dimensional list and returns a one-dimensional list containing all the unique entries in the list. The entries in the list returned do not need to be ordered in any particular way. For example, in the list [[1, 0, 1], [0, 1, 0]] there are only two unique values (0 and 1) so the function would return the list [0, 1] (or the list [1, 0] -- both are valid). The list passed as a parameter should not be altered by the function

#Method definition for the function unique()

#The method takes in one argument which is a two dimensional list

def unique (twoDList) :

   #Create an empty one dimensional list

   oneDList = [];

   #Convert the two dimensional list to a one dimensional list by

   #looping through the two dimensional list and putting each element in

   # the one dimensional list using the python in-built append() function

   for i in range(len(twoDList)) :  

        for j in range(len(twoDList[i])) :  

           oneDList.append(twoDList[i][j])

   #Convert the one dimensional list to a set to make the list unique

   #by using the python in-built set() function

   mySet = set(oneDList)

   #Convert the unique list back to a list

   uniqueList = list(mySet)

   #Return the unique list

   return uniqueList

#End of method declaration

#Call the method with an arbitrary two dimensional list, and print the result

print(unique([[3,5,3],[4,6,3],[1,7,8]]))

print(unique([[1, 0, 1], [0, 1, 0]]))

[1, 3, 4, 5, 6, 7, 8]

[0,1]

The above code has been written in Python. It contains comments explaining each line of the code. Please go through the comment. The actual code has been written in bold-face to differentiate it from comments. Note that in Python, indentation really matters. Therefore, the code must be written with the appropriate indentations like those shown in the code above.

For clarity, the code is re-written as follows without comments.

def unique (twoDList) :

   oneDList = [];

   for i in range(len(twoDList)) :  

        for j in range(len(twoDList[i])) :  

           oneDList.append(twoDList[i][j])

   mySet = set(oneDList)

   uniqueList = list(mySet)

   return uniqueList

print(unique([[3,5,3],[4,6,3],[1,7,8]]))

print(unique([[1, 0, 1], [0, 1, 0]]))

the answer to your question would be something called hertz. the unit of measurement called a hertz, which is one cycle per second, is used to measure clock speed. within one cycle, your computer processes, refreshes, or updates itself to the changes that it is going through. in the case of computer clock speed, one hertz equals one tick per second. the clock speed of computers is usually measured in megahertz (mhz) or gigahertz (ghz)

the benefits of raising bc’s minimum wage to $15 through a

series of pre-announced staged increases far outweigh the likely

costs. claims that such an increase will lead to massive job losses

in low-wage sectors of the economy are not credible.

• an increase to $15 would result in some job losses. however,

the likely impact is less than a 1 per cent reduction in the overall

provincial employment rate, and a 7.6 per cent reduction in the

employment rate for those currently making between $10.25

and $15 per hour.

• the large job loss impacts predicted by some opponents of

minimum wages misrepresent the existing economic research. in

reading and using that research, it is important to recognize that

estimates showing job loss effects of minimum wage increases

apply only to teenagers. estimated effects for young adult and

adult workers range from insignificant to non-existent.

• focusing on teenage workers plays up an inaccurate stereotype

about who earns very low wages in bc and leads to exaggerated

claims about job losses. only 26 per cent of those currently making

between $10.25 and $12 are teenagers, and only 5 per cent

of those making between $12 and $15 are teenagers.