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Creating emails while accounting for two first names when using input()

My code below works as long as all of the student's name consists of only two names - ex. Julie Andrews. But, when generating the student's emails, I'm trying to account for the students who have two first names - ex. Mary Jane Stewart. I want it to output something like MJStewart123#gmail.org, vs. what my current code will print, which is MJane123#gmail.com - totally ignoring the student's last name.
After hours of researching Google, I have tried updating my create_emails fx to change my original variable first_last = name.split(" ") to something like first, middle, last = name.split(" ") or first_last = name.split(" ", 2) while also, respectively, updating the line utilizing the attribute .append from its original to student_emails.append(first_last[0][0] + first_last1 + first_last[2]+ last_three_sid + "#gmail.com") or student_emails.append(first[0] + middle[0] + last + last_three_sid + "#gmail.com"). All attempts have obviously returned some form of an error...
The attached Stack Overflow article is the closest thing I could find whose logic might be applicable to what I'm trying to accomplish here, specifically the comment by Manfred, but in reading it, I don't know how to apply what they've done to my program... because I don't quite understand what it is that I'm reading... since I'm such a newbie at all this. I'd appreciate any help you can offer.
student_names = []
def create_names():
count = 1
while count <= 5:
name = input("Enter student name, please. ")
student_names.append(name)
count += 1
create_names()
import random
student_ids = []
def create_ids():
student_id = random.randint(111111,999999)
return student_id
def create_id_list():
for name in student_names:
student_ids.append(create_ids())
create_id_list()
student_emails = []
def create_emails():
for name in student_names:
first_last = name.split(" ")
sid = str(student_ids[student_names.index(name)])
len_sid = len(sid)
last_three_sid = sid[len_sid-3:len_sid]
student_emails.append(first_last[0][0] + first_last[1] + last_three_sid + "#gmail.com") #ignores last index if one is provided.
create_emails()
def student_info():
for name in student_names:
name_pos = student_names.index(name)
print("\n" + "name: " + name)
print("id: " + str(student_ids[name_pos]))
print("email: " + student_emails[name_pos])
student_info()
Finding and first and middle initials in a list of names in python

You could do it like this I guess:
student_names = []
def create_names():
count = 1
while count <= 5:
name = input("Enter student name, please. ")
student_names.append(name)
count += 1
create_names()
import random
student_ids = []
def create_ids():
student_id = random.randint(111111,999999)
return student_id
def create_id_list():
for name in student_names:
student_ids.append(create_ids())
create_id_list()
student_emails = []
def create_emails():
for name in student_names:
email_name = ""
first_last = name.split(" ")
for i, v in enumerate(first_last):
if i > len(first_last)-2:
break
email_name+=v[0]
email_name = email_name+first_last[-1]
sid = str(student_ids[student_names.index(name)])
len_sid = len(sid)
last_three_sid = sid[len_sid-3:len_sid]
student_emails.append(email_name + last_three_sid + "#gmail.com") #ignores last index if one is provided.
create_emails()
def student_info():
for name in student_names:
name_pos = student_names.index(name)
print("\n" + "name: " + name)
print("id: " + str(student_ids[name_pos]))
print("email: " + student_emails[name_pos])
student_info()
Result:
Enter student name, please. Julie Andrews
Enter student name, please. Mary Jane Stewart
Enter student name, please. Jack Hendricks
Enter student name, please. Maria Basset Juliett
Enter student name, please. Marco Hansen
name: Julie Andrews
id: 742536
email: JAndrews536#gmail.com
name: Mary Jane Stewart
id: 823274
email: MJStewart274#gmail.com
name: Jack Hendricks
id: 590875
email: JHendricks875#gmail.com
name: Maria Basset Juliett
id: 982168
email: MBJuliett168#gmail.com
name: Marco Hansen
id: 671240
email: MHansen240#gmail.com

The code is too redundant and could be implemented with some simpler structures.
name construction can use split combined with join function
storage can also use dictionaries to store multiple information about the same object, compared to multiple lists so more concise and efficient
The code itself is not a big problem, mainly the syntax structure requires more skilled,let's encourage each other in our endeavours
try this:
import random
info = []
# You can use for loop if you already know how many times it will loop
for i in range(5):
name = input("Enter student name, please:") # get name
student_id = random.randint(111111, 999999) # get id
name_structure = name.split()
email = "{name_abbr}{last_name}{sid}#gmail.com".format(
name_abbr="".join([item[0] for item in name_structure[:-1]]), # Generate initials
last_name=name_structure[-1], # Generate the last part of the name
sid=str(student_id)[-3:] # Generate the id in the mailbox
)
info.append({"name": name, "id": student_id, "email": email}) # Store to the list, or print directly
# Print Information
for item in info:
print("name:", item["name"])
print("id:", item["id"])
print("email:", item["email"])
print()

Another question: Phone dictionary problem 'while-loop' using Error

Simply question making phone dictionary
What I want to do is putting person's name and number and finding them in dictionary!
Examples what I want to do
Enter command (a, f, d, or q).: a
Enter new name................: Perry
Enter new phone number........: 229-449-9683
Enter command (a, f, d, or q).: f
Enter name to look up...:
I would like to find full name and number when I type
Phone dictionary code what I wrote so far:
phone_dict = {}
command = input('Enter command (a, f, d, or q).: ')
newname = input('Enter new name................: ')
newphone = input('Enter new phone number........: ')
while True:
if command == 'a':
newname
newphone
phone_dict[newname] = newphone
print(phone_dict)
# In here, 'while-loop' does not work.
In there, if I enter 'a' command, and type the name
The dictionary is supposed to be { Perry: 229-449-9683}
Thanks, The question might be little confused, but if you can help this out, I am very happy!

To find the result from the dictionary, you can loop through the items and check if the key contains the string you want to find. If you want to get all values which satisfy your query, you can create another list or dictionary and store the items you find:
phone_dict = {
"Han Perry": "1234",
"Harry Gildong": "2345",
"Hanny Test": "123",
}
find_str = "Han"
result = {}
for key, value in phone_dict.items():
# Converting it to lower makes it case insensitive
if find_str.lower().strip() in key.lower():
result[key] = value
print(result)
# {'Han Perry': '1234', 'Hanny Test': '123'}
Take note that this will run through all of the values of the dictionary: O(n)

To find the number using the first o last name of the person you could do:
a = 'Add a new phone number'
d = 'Delete a phone number'
f = 'Find a phone number'
q = 'Quit'
phone_dict = {}
while True:
# Gets the user command every loop
command = input('Enter command (a, f, d, or q).: ')
# Add a new registry to the directory
if command == 'a':
newname = input('Enter new name................: ')
newphone = input('Enter new phone number........: ')
phone_dict[newname] = newphone
print(phone_dict)
# Find a registry on the directory
elif command == "f"
query = input("Enter name to look up...: ")
match = None
for key in phone_dict.keys():
if query.strip() in key:
match = phone_dict[key]
break
if match is None:
print(f"The name {query} could not be found on the directory")
else:
print(f"The phone number of {query} is {match}")
elif command == "d":
# Delete registry
elif command == "q":
# Quits program
else:
print(f"The command {command} was not found, please try again!")
In this case, I am using query.strip() to remove any extra start/end spaces that could cause to not find the person.

how to split string from a integer in python?

My input needs to be for example
name surname, street name 22b, 10000 Zagreb
and it need to output like this:
name and surname: name surname
Street: street name
street number: 22
house number: b
postal code: 100000
place: zagreb
and this is my code
whole_string =input("Person: ")
string_list = whole_string.split(", ")
split_street_and_street_number = string_list[1].split(" ")
postal_code_and_city = string_list[2].split(" ")
print(f"name and surname: {string_list[0]}")
print(f"street: {split_street_and_street_number[0]}")
print(f"street number: {split_street_and_street_number[1]}")
print(f"postal code: {postal_code_and_city[0]}")
print(f"city: {postal_code_and_city[1]}")

Please check out this.
import re
text = '22b'
street_number =" ".join(re.findall("[0-9]+", text))
house_number =" ".join(re.findall("[a-zA-Z]+", text))
print(street_number)
print(house_number)

You can find the index of the first letter and split the string using that index:
def find_index_of_first_letter(text):
for index, value in enumerate(text):
try:
int(value)
except ValueError:
return index
print('No letter in the text')
text = '22b'
first_letter = find_index_of_first_letter(text)
number, letters = text[:first_letter], text[first_letter:]

How do I access the array using the for loop in the while loop

I am working with an external file which has data in the form of:
-12345 CSEE 35000 Bart Simpson
-12346 CSEE 25000 Harry Potter
-12350 Economics 30000 Krusty The Clown
-13123 Economics 55000 David Cameron
With the first item being the ID, the second the subject, the third the salary, and the rest being the name of the person.
In part of my program I am trying to print the information of the people who have salaries between values submitted by the user. I have put all the data in a list called lecturers then I put all the salaries in a separate list called lecturers salary and tried to make them integers because at first I thought the reason the for loop wasn't working was because when trying to access them from the lectures loop I thought they might still be part of a string at this point.
I have already used a loop in my program to print all the people who teach a specific subject. This subject is submitted by the user. I tried to use a for loop again for the salaries but its not working.
print""
# To God be the Glory
lecturer = []
lecturer_salary = []
x = 0
a = " "
print ""
String = raw_input("Please enter the lecturers details: ")
print ""
def printFormat(String):
String = String.split()
lastname = String[-1]
firstnames = " ".join(String[3:-1])
name = ", ".join([lastname, firstnames])
ID_Subject = " ".join(String[0:2])
money = String[2]
print "%s,%s %s %s" % (lastname,firstnames,ID_Subject,money)
printFormat(String)
while x < len(lecturer):
lecturer_salary.append(int(lecturer [x][2]))
x = x + 1
print ""
try:
fname = input("Enter filename within " ": ")
with open(fname) as f:
for line in f:
data = line.split()
printFormat(line)
line = line.split()
lecturer.append(line)
except IOError as e :
print("Problem opening file")
print ""
print ""
answer = raw_input("Would you like to display the details of lectureers from a particular department please enter YES or NO: ")
if answer == "YES" :
print ""
department = raw_input("Please enter the department: ")
print ""
while x < len(lecturer) :
for line in lecturer:
if lecturer[x][1] == department:
a = lecturer[x]
a = ' '.join(a)
printFormat(a)
x = x + 1
**elif answer == "NO" :
print ""
answer2 = raw_input ("Would you like to know all the lecturers within a particular salary range: ")
print ""
if answer2 == "YES":
lower_bound = int(input("Please enter the lower bound of the salary range: "))
upper_bound = int(input("Please enter the upper bound of the salary range: "))
print ""
while x < len(lecturer) :
for line in lecturer_salary:
if lower_bound < lecturer_salary[x] < upper_bound :
print lecturer_salary[x]
x = x + 1**
else:
print ""
print "Please enter a valid input"

So, you have an array of lecturer and one of lecturer salary. the
for line in lecturer_salary:
is not needed - just the while followed by the if. Note that this will only print out the salary, not the lecturer details. Since x is the index to both arrays you can access lecturer[x] for the rest. In truth you don't need the lecturer_salary at all, just walk through lecturer and check:
while x < len(lecturer) :
if lower_bound < lecturer[x][2] < upper_bound :
a = lecturer[x]
a = ' '.join(a)
printFormat(a)
x = x + 1

For starters, you shouldn't name your variable with a capital letter like String or Id_Subject.
It is simpler to break code into functions and try using a dictionary or class to improve readability and extensibility.
Here is a minimal code using class:
lecturers = [] # To store Lecturer instances, which isn't necessary
class Lecturer():
def __init__(self, id, subject, salary, name):
self.id = id
self.subject = subject
self.salary = salary
self.name = name
def readfile(filename):
"""read each line in a file and yield a list of fields"""
with open(filename, "r") as f:
for line in f.readlines():
# return a list of fields
yield line.replace("\n", "").split()
def new_lecturer(detail):
"""Return a new lecturer instance from a list of fields"""
return Lecturer(detail[0],
detail[1],
detail[2],
{"firstname": detail[3],
"lastname": detail[4]
})
def print_lecturer_detail(lecturer):
"""Accept a lecturer instance and print out information"""
print "{0},{1} {2} {3}".format(lecturer.name["lastname"],
lecturer.name["firstname"],
lecturer.id,
lecturer.salary)
def main():
"""This is where all the main user interaction should be"""
fname = raw_input("Enter filename: ")
for lecturer in (readfile(fname)):
lecturers.append(new_lecturer(lecturer))
print ""
answer = raw_input("Would you like to display lecturers by department(Y/N)?: ")
if answer == "Y":
print ""
department = raw_input("Please enter the department: ")
print ""
for lecturer in lecturers:
if lecturer.subject == department:
print_lecturer_detail(lecturer)
elif answer == "N":
# implement salary code here
pass
if __name__ == '__main__':
main()
This may be an overkill now, but it's better than dealing with lists in a long run. You'll see that dealing with properties become much simpler. You may want to improve each function further and make it more modular and reusable.
#Paul Morrington has the straight answer on the while part.

Grouping string input by count

I'm trying to do a question out of my book and it asks:
Implement function names that takes no input and repeatedly asks the
user to enter a student's first name. When the user enters a blank
string, the function should print for every name, the number of
students with that name.
Example usage:
Usage:
names()
Enter next name: Valerie
Enter next name: Bob
Enter next name: Valerie
Enter next name: John
Enter next name: Amelia
Enter next name: Bob
Enter next name:
There is 1 student named Amelia
There are 2 students named Bob
There is 1 student named John
There are 2 students named Valerie
So far I have this code:
def names():
names = []
namecount = {a:name.count(a) for a in names}
while input != (''):
name = input('Enter next name: ')
names = name
if input == ('')
for x in names.split():
print ('There is', x ,'named', names[x])
I'm really lost here and any input would help out tons. Also if possible please explain how to fix my code

There are a lot of issues with namings in your function, you are using such variables like 'names' that is used for function name as well as 'input' that is a python function name for reading user input - so you have to avoid using this. Also you defining a namecount variable as a dict and trying to initialize it before fill. So try to check solution below:
def myFunc():
names = []
name = ''
while True: #bad stuff you can think on your own condition
name = raw_input('press space(or Q) to exit or enter next name: ')
if name.strip() in ('', 'q', 'Q'):
for x in set(names):
print '{0} is mentioned {1} times'.format(x, names.count(x))
break
else:
names.append(name)
myFunc()
OR:
from collections import defaultdict
def myFunc():
names = defaultdict(int)
name = ''
while True: #bad stuff you can think on your own condition
name = raw_input('press space(or Q) to exit or enter next name: ')
if name.strip() in ('', 'q', 'Q'):
for x in set(names):
print '{0} is mentioned {1} times'.format(x, names[x])
break
else:
names[name] += 1

I rewrote your function for you:
def names():
names = {} # Creates an empty dictionary called names
name = 'cabbage' # Creates a variable, name, so when we do our while loop,
# it won't immediately break
# It can be anything really. I just like to use cabbage
while name != '': # While name is not an empty string
name = input('Enter a name! ') # We get an input
if name in names: # Checks to see if the name is already in the dictionary
names[name] += 1 # Adds one to the value
else: # Otherwise
names[name] = 1 # We add a new key/value to the dictionary
del names[''] # Deleted the key '' from the dictionary
for i in names: # For every key in the dictionary
if names[i] > 1: # Checks to see if the value is greater for 1. Just for the grammar :D
print("There are", names[i], "students named", i) # Prints your expected output
else: # This runs if the value is 1
print("There is", names[i], "student named", i) # Prints your expected output
When doing names():
Enter a name! bob
Enter a name! bill
Enter a name! ben
Enter a name! bob
Enter a name! bill
Enter a name! bob
Enter a name!
There are 3 students named bob
There are 2 students named bill
There is 1 student named ben

Let's analyse your code:
def names():
names = []
namecount = {a:name.count(a) for a in names}
while input != (''):
name = input('Enter next name: ')
names = name
if input == ('')
for x in names.split():
print ('There is', x ,'named', names[x])
There seem to be a few problems, let's list them
The while loop's conditional
What you want to do check if input from user is '' (nothing)..
input is a built-in function for getting input from user, so it never will be ('').
The names = name statement
What you want to do is add name to the list names.
Here you are changing names to a string, which isn't what you want.
The if's conditional
same as 1.
The for loop
let's ignore.. just not valid.. here..
We fix these problems as follows(solution has same numbering as problem above that it solves)
Change the conditional to something like name != ''.
Also, before the loop begins, you need to get input once for this to work, which in this case has a bonus, the first input can have a different prompt.
Use names.append(name) to add name to names.
Same as 1.
Just look at the for loop below...
Try this
def names():
names = []
name = input('Enter a name: ').strip() # get first name
while name != '':
names.append(name)
name = raw_input('Enter next name: ').strip() # get next name
for n in set(names): # in a set, no values are repeated
print '%s is mentioned %s times' % (n, names.count(n)) # print output

def names():
counters = {}
while True:
name = input('Enter next name:')
if name == ' ':
break
if name in counters:
counters[name] += 1
else:
counters[name] = 1
for name in counters:
if counters[name] == 1:
print('There is {} student named {}'.format(counters[name],name))
else:
print('There are {} student named {}'.format(counters[name],name))
names()