Python Using While loop avoid duplicate print

You can assign a second variable (y) to the original variable (x) and then assign a new value to x. Then, in an if statement check if they aren't equal (!=).

The code:

def parse():
    x = ''
    while True:
        y = x
        x = driver.find_element_by_xpath('//*[@id="bets-history"]/li[0]').text
        if y != x:
            print(x)

data = ['1.54x','1.54x','1.54x','1.54x','1.54x','1.54x','1.54x','1.54x','13.5x','13.5x','13.5x','13.5x','13.5x','13.5x','13.5x','13.5x']

previous_value = ''
for value in data:
    if value != previous_value:
        print(value)
        previous_value = value

As you only want every value to be printed once, you can define a list, xs.

For each x defined, if the value of the x is already in the xs list, don't print or append the value of x to xs; else, print the value of x and append the value of x to xs:

def parse():
    xs = []
    while True:
        x = driver.find_element_by_xpath('//*[@id="bets-history"]/li[0]').text
        if x not in xs:
            xs.append(x)
            print(x)

You have to store the uniquely received values. Later you can check if the newly received value already exists then do not print this value otherwise print it to console.

def parse():
    stored_data = []
    while True:
        x = driver.find_element_by_xpath('//*[@id="bets-history"]/li[0]').text
        if x not in stored_data:
            print(x)
            stored_data.append(x)

Use a set for faster lookups instead of list.

Since in your question there are only two values, it won't be much of a difference looking up in list and a set. But if there are n values,

list will take O(n) time to lookup

set will take just O(1) time to lookup

def parse():
    s = set()
    while True:
        x = driver.find_element_by_xpath('//*[@id="bets-history"]/li[0]').text
        if x not in s:
           s.add(x)
           print(x)