Develop Reference

Python: how prevent duplication of randomly chosen key/value pair

def fill2_line1():
x2 = random.choice(list(twoSyllValues.items()))
line1.append(x2)
magicList = ([x[1] for x in line1])
if 1 in magicList:
fill2_line1()
fill2_line1()
complete_line = line1[0][0] + " " + line1[1][0] + " " +
line1[2][0]
print(complete_line)
This is the part in question - the whole program is over 150 lines. It works perfectly as is, but sometimes I'm getting duplicated words. To refine the code, I want to make sure that the key in the first random tuple selection is not duplicated in the next two selected tuples IN that list, line1. The only way I can think of is having another dictionary created after the first word selection and then exclude that key,value pair from new dictionary and then make the next two random.choice calls. I feel like there must be a simpler way, but I just don't have the experience yet.

I can think of three ways to solve the problem you are facing. I will write a sample code that you can refer to since I do not know your whole code.
Delete the key before the second random.choice calls and adding them back in if you need them. This is somewhat similar to creating another dictionary but more efficient.
def fill2_line1():
x2 = random.choice(list(twoSyllValues.items()))
line1.append(x2)
return x2
magicList = ([x[1] for x in line1])
if 1 in magicList:
deleted_list = []
# run first random.choice
deleted_item = fill2_line1()
deleted_list.append(deleted_item)
del twoSyllValues[deleted_item[0]]
# run second random.choice
deleted_item = fill2_line1()
deleted_list.append(deleted_item)
del twoSyllValues[deleted_item[0]]
complete_line = line1[0][0] + " " + line1[1][0] + " "
+ line1[2][0]
print(complete_line)
# add the deleted items back in after work is done
for k, v in deleted_list:
twoSyllValues[k] = v
Run random.choice until there is no duplicate item.
def fill2_line1(deleted_set):
x2 = random.choice(list(twoSyllValues.items()))
while x2 not in deleted_set:
x2 = random.choice(list(twoSyllValues.items()))
line1.append(x2)
deleted_set.add(x2)
magicList = ([x[1] for x in line1])
if 1 in magicList:
deleted_set = set([])
fill2_line1(deleted_set)
fill2_line1(deleted_set)
complete_line = line1[0][0] + " " + line1[1][0] + " "
+ line1[2][0]
print(complete_line)
Keep another list just for the random.choice
def fill2_line1(items_list):
x2 = random.choice(items_list)
line1.append(x2)
items_list.remove(x2)
magicList = ([x[1] for x in line1])
if 1 in magicList:
items_list = list(twoSyllValues.items())
fill2_line1(items_list)
fill2_line1(items_list)
complete_line = line1[0][0] + " " + line1[1][0] + " "
+ line1[2][0]
print(complete_line)
Hope my answer gave you some ideas!

I'm trying to WAP to convert denary to binary without using list

I have already seen a couple of programs for this. Just wanted to know what's wrong with this logic. It keeps on returning 'str' object does not support item assignment. I looked this up but still couldn't find a reason this occurs.I'm just a newbie so apologies in advance if I'm just overthinking things.
x = dec
a = 5
n = 1
remainder = str()
binary = str()
while a != 1:
a = x // 2
b = x % 2
x = a
z = str(b)
remainder = str(remainder + z)
if a == 1:
b = 1
z = str(b)
remainder = str(remainder + z)
print(remainder)
asd = len(remainder)
for y in range(1, asd + 1):
binary[y:y + 1] = remainder[-y:-y - 1]
print("It's binary form is ", binary)

The problem is here:
for y in range(1, asd + 1):
binary[y:y + 1] = remainder[-y:-y - 1]
# ^^^^^^^^^^^^^ error
What do you expect this to do? Python strings are immutable: you cannot assign to a string slice. You can only use the value (right-hand side of an assignment, for example), or replace the variable's entire value.
I suspect that you're trying to reverse the string. Do it like this:
rev = remainder[::-1] # Whole string, but in reverse order

Check result using 4 operations based with Python

I'm struggling to make a Python program that can solve riddles such as:
get 23 using [1,2,3,4] and the 4 basic operations however you'd like.
I expect the program to output something such as
# 23 reached by 4*(2*3)-1
So far I've come up with the following approach as reduce input list by 1 item by checking every possible 2-combo that can be picked and every possible result you can get to.
With [1,2,3,4] you can pick:
[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]
With x and y you can get to:
(x+y),(x-y),(y-x),(x*y),(x/y),(y/x)
Then I'd store the operation computed so far in a variable, and run the 'reducing' function again onto every result it has returned, until the arrays are just 2 items long: then I can just run the x,y -> possible outcomes function.
My problem is this "recursive" approach isn't working at all, because my function ends as soon as I return an array.
If I input [1,2,3,4] I'd get
[(1+2),3,4] -> [3,3,4]
[(3+3),4] -> [6,4]
# [10,2,-2,24,1.5,0.6666666666666666]
My code so far:
from collections import Counter
def genOutputs(x,y,op=None):
results = []
if op == None:
op = str(y)
else:
op = "("+str(op)+")"
ops = ['+','-','*','/','rev/','rev-']
z = 0
#will do every operation to x and y now.
#op stores the last computated bit (of other functions)
while z < len(ops):
if z == 4:
try:
results.append(eval(str(y) + "/" + str(x)))
#yield eval(str(y) + "/" + str(x)), op + "/" + str(x)
except:
continue
elif z == 5:
results.append(eval(str(y) + "-" + str(x)))
#yield eval(str(y) + "-" + str(x)), op + "-" + str(x)
else:
try:
results.append(eval(str(x) + ops[z] + str(y)))
#yield eval(str(x) + ops[z] + str(y)), str(x) + ops[z] + op
except:
continue
z = z+1
return results
def pickTwo(array):
#returns an array with every 2-combo
#from input array
vomit = []
a,b = 0,1
while a < (len(array)-1):
choice = [array[a],array[b]]
vomit.append((choice,list((Counter(array) - Counter(choice)).elements())))
if b < (len(array)-1):
b = b+1
else:
b = a+2
a = a+1
return vomit
def reduceArray(array):
if len(array) == 2:
print("final",array)
return genOutputs(array[0],array[1])
else:
choices = pickTwo(array)
print(choices)
for choice in choices:
opsofchoices = genOutputs(choice[0][0],choice[0][1])
for each in opsofchoices:
newarray = list([each] + choice[1])
print(newarray)
return reduceArray(newarray)
reduceArray([1,2,3,4])

The largest issues when dealing with problems like this is handling operator precedence and parenthesis placement to produce every possible number from a given set. The easiest way to do this is to handle operations on a stack corresponding to the reverse polish notation of the infix notation. Once you do this, you can draw numbers and/or operations recursively until all n numbers and n-1 operations have been exhausted, and store the result. The below code generates all possible permutations of numbers (without replacement), operators (with replacement), and parentheses placement to generate every possible value. Note that this is highly inefficient since operators such as addition / multiplication commute so a + b equals b + a, so only one is necessary. Similarly by the associative property a + (b + c) equals (a + b) + c, but the below algorithm is meant to be a simple example, and as such does not make such optimizations.
def expr_perm(values, operations="+-*/", stack=[]):
solution = []
if len(stack) > 1:
for op in operations:
new_stack = list(stack)
new_stack.append("(" + new_stack.pop() + op + new_stack.pop() + ")")
solution += expr_perm(values, operations, new_stack)
if values:
for i, val in enumerate(values):
new_values = values[:i] + values[i+1:]
solution += expr_perm(new_values, operations, stack + [str(val)])
elif len(stack) == 1:
return stack
return solution
Usage:
result = expr_perm([4,5,6])
print("\n".join(result))

Alternating sums of a list [duplicate]

This question already has answers here:
Writing a function that alternates plus and minus signs between list indices
(7 answers)
Closed 2 years ago.
9.Write a program that accepts 9 integers from the user and stores them in a list. Next, compute the alternating sum of all of the elements in the list. For example, if the user enters
1 4 9 16 9 7 4 9 11
then it computes
1 – 4 + 9 – 16 + 9 – 7 + 4 – 9 + 11 = –2
myList = []
value = None
count = 0
while count != 9:
value = int(input("Please enter a number: "))
myList.append(value)
count = count + 1
if count == 9:
break
print(myList)
def newList(mylist):
return myList[0] - myList[1] + myList[2] - myList[3] + myList[4] - myList[5] + myList[6] - myList[7] + myList[8]
x = newList(myList)
print(x)
My code returns the correct answer, but I need it to print out the actual alternating sums as in the example. I have been stuck on this for a while. I am having a mental block on this and havent been able to find anything similar to this online.
I appreciate any help or tips.
Also, this is python 3.
Thank you.

a=[1, 4, 9, 16, 9, 7, 4, 9, 11]
start1=0
start2=1
sum1=0
first_list=[a[i] for i in range(start1,len(a),2)]
second_list=[a[i] for i in range(start2,len(a),2)]
string=''
for i,j in zip(first_list,second_list):
string+=str(i)+'-'+str(j)+'+'
string.rstrip('+')
print('{}={}'.format(string,str(sum(first_list)-sum(second_list))))
Output
1-4+9-16+9-7+4-9+=-2

Try doing this:
positives = myList[::2]
negatives = myList[1::2]
result = sum(positives) - sum(negatives)
print ("%s = %d" % (" + ".join(["%d - %d" % (p, n) for p, n in zip(positives, negatives)]), result))
I'll explain what I'm doing here. The first two lines are taking slices of your list. I take every other number in myList starting from 0 for positives and starting from 1 for negatives. From there, finding the result of the alternating sum is just a matter of taking the sum of positives and subtracting the sum of negatives from it.
The final line is somewhat busy. Here I zip positives and negatives together which produces a list of 2-tuples where of the form (positive, negative) and then I use string formatting to produce the p - n form. From there I use join to join these together with the plus sign, which produces p0 - n0 + p1 - n1 + p2 - n2.... Finally, I use string formatting again to get it in the form of p0 - n0 + p1 - n1 + p2 - n2 ... = result.

You can do as you did but place it in a print statement
print(myList[0] + " - " + myList[1] + " + " + myList[2] + " - " + myList[3] + " + " + myList[4] + " - " + myList[5] + " + " + myList[6] + " - " + myList[7] + " + " + myList[8] + " = " + x)
Its not perfectly clean, but it follows your logic, so your teacher won't know you got your solution from someone else.

Something along the lines of the following would work:
def sumList(theList):
value = 0
count = 0
steps = ""
for i in theList:
if count % 2 == 0:
value += i
steps += " + " + str(i)
else:
value -= i
steps += " - " + str(i)
count += 1
print(steps[3:])
return value
print(sumList(myList))
It alternates between + and - by keeping track of the place in the list and using the modulus operator. Then it calculates the value and appends to a string to show the steps which were taken.

You can also do something like below once your 9 or more numbers list is ready
st = ''
sum = 0
for i, v in enumerate(myList):
if i == 0:
st += str(v)
sum += v
elif i % 2 == 0:
st += "+" + str(v)
sum += v
else:
st += "-" + str(v)
sum -= v
print("%s=%d" % (st, sum))
It prints : 1-4+9-16+9-7+4-9+11=-2

Create multiplication table?

Am a beginner in Programming and am practicing how to use nested for loops to make a multiplication table in python 2.7.5.
Here is my code
x=range(1,11)
y=range(1,11)
for i in x:
for j in y:
print i*j
pass
well,the result is correct but it does not appear in a square matrix form as i wish.Please help me improve the code

You should print without a line break.
x = range(1,11)
y = range(1,11)
for i in x:
for j in y:
print i*j, # will not break the line
print # will break the line

you may add formatting to keep constant cell width
x = range(1,11)
y = range(1,11)
for i in x:
for j in y:
# substitute value for brackets
# force 4 characters, n stands for number
print '{:4n}'.format(i*j), # comma prevents line break
print # print empty line

Python's print statement adds new line character by default to the numbers you wish to have in your output. I guess you would like to have just a trailing spaces for inner loop and a new line character at the end of the outer loop.
You can achieve this by using
print i * j, # note the comma at the end (!)
and adding just a new line at the end of outer loop block:
print ''
To learn more about the trailing coma, and why it works, look here: "How to print in Python without newline or space?". Mind that it works differently in Python 3.
The final code should look like:
x=range(1,11)
y=range(1,11)
for i in x:
for j in y:
print i*j,
print ''
You can also look for '\t' special character which would allow you to get better formatting (even this old resource is good enough: https://docs.python.org/2.0/ref/strings.html)

USE This Code. It works MUCH better. I had to do this for school, and I can tell you that after putting about 4 hours into this it works flawlessly.
def returnValue(int1, int2):
return int1*int2
startingPoint = input("Hello! Please enter an integer: ")
endingPoint = input("Hello! Please enter a second integer: ")
int1 = int(startingPoint)
int2 = int(endingPoint)
spacing = "\t"
print("\n\n\n")
if int1 == int2:
print("Your integers cannot be the same number. Try again. ")
if int1 > int2:
print("The second number you entered has to be greater than the first. Try again. ")
for column in range(int1, int2+1, 1): #list through the rows(top to bottom)
if column == int1:
for y in range(int1-1,int2+1):
if y == int1-1:
print("", end=" \t")
else:
individualSpacing = len(str(returnValue(column, y)))
print(y, " ", end=" \t")
print()
print(column, end=spacing)
for row in range(int1, int2+1, 1): #list through each row's value. (Go through the columns)
#print("second range val: {:}".format(row))
individualMultiple = returnValue(row, column)
print(individualMultiple, " ", end = "\t")
print("")
Have a good day.

#Generate multiplication table by html
import random
from copy import deepcopy
N = 15
colors = ['F','E','D','C','B','A']
i = 0
colorsall = []
while i < N:
colornow = deepcopy(colors)
random.shuffle(colornow)
colornow = "#"+"".join(colornow)
colorsall.append(colornow)
i += 1
t = ""
for i in range(1,N+1):
s = ''
for j in range(1,N+1):
if j >= i:
s += '<td style="background-color:' + colorsall[i-1] + '">'+str(i*j)+'</td>'
else:
s += '<td style="background-color:' + colorsall[j-1] + '">'+str(i*j)+'</td>'
s = "<tr>" + s + "</tr>"
t = t + s + '\n'
print('<table>' + t + '</table>')
Taken from https://todaymylearn.blogspot.com/2022/02/multiplication-table-in-html-with-python.html [Disclosure : my blog]