'A'[ ]Note that the first index in a sequence is always zero.
Since the first index of a sequence is zero, the last index is going to be the length of the string minus 1
four_letter_anagramstring, and four integers a, b, c and da, b, c and d – concatenate the individual charactersfour_letter_anagram("balm", 2, 1, 3, 0) should return lambAdd the following test cases to the code below:
loin returns lionlugs returns slugreap returns pearSubmit your four_letter_anagram function to Gradescope for attendance.
Name your file anagram.py
in operatorThe in operator determines whether a given value is a constituent element of a sequence (such as a string)
Remember the string method.isnumeric()?
is numeric for one characterWrite a Python function that does the following:
is_numeric_one_charstring argument of length of 1 (check if argument is valid with len())True if the argument is a digit (0-9), False otherwise (use the in operator)is_numeric() for any string lengthis_numericstring argument of any lengthTrue if the first character of the argument is a digit (0-9), False otherwise (remember to use [ ] to index the string and the in operator)Test cases:
is_numeric()What if we want to check every character in a string of any length?
What if the string is of length 2 or 5 or 45?
is_numeric()is_numericstring argument of any lengthTrue if every character in the argument is a digit (0-9), False otherwise[ ] to index the string, use a while loop with a len() condition, and an index – something like string[index] with index being updated inside a while loopis_numeric() solutiondef is_numeric(my_string):
index = 0
while index < len(my_string):
if my_string[index] not in "0123456789":
return False
index += 1
return True
def main():
print( is_numeric("234") ) # True
print( is_numeric("abc") ) # False
print( is_numeric("12c") ) # False
print( is_numeric("12.3") ) # False
main()True
False
False
Falseis_numeric() solutionHow do we change our function so that the last test case (“12.3”) returns True instead?
def is_numeric(my_string):
index = 0
while index < len(my_string):
if my_string[index] not in "0123456789":
return False
index += 1
return True
def main():
print( is_numeric("234") ) # True
print( is_numeric("abc") ) # False
print( is_numeric("12c") ) # False
print( is_numeric("12.3") ) # False
main()True
False
False
Falseis_numeric() solutionHow do we change our function so that the last test case (“12.3”) returns True instead?
def is_numeric(my_string):
index = 0
while index < len(my_string):
if my_string[index] not in "0123456789.":
return False
index += 1
return True
def main():
print( is_numeric("234") ) # True
print( is_numeric("abc") ) # False
print( is_numeric("12c") ) # False
print( is_numeric("12.3") ) # True
print( is_numeric("1.2.3") ) # True
main()True
False
False
True
Trueis_numeric() solutiondef is_numeric(my_string):
# create control variable for one decimal point
decimal_point = False
# create index variable
index = 0
while index < len(my_string):
# first check, can character be found in a number?
if my_string[index] not in "0123456789.":
return False
# second check, if a period, has a period been found before?
if my_string[index] == ".":
if decimal_point: # a previous period was detected
return False
else: # first period detected
decimal_point = True
# increment index
index += 1
# while loop executed without returning False
# that means every character is valid, so return True
return True
def main():
print( is_numeric("234") ) # True
print( is_numeric("abc") ) # False
print( is_numeric("12c") ) # False
print( is_numeric("12.3") ) # True
print( is_numeric("1.2.3") ) # False
main()True
False
False
True
FalseYou have 10 minutes to complete the quiz
main(), no need to print test casesBuilt-in functions you can use: round(), input(), float(), str(), int() — you don’t have to use all of these