No numerals, easier way to solve it on Python?

Question

I solved the No numerals problem, but i know that there must be a better way to solve it:

I solved it replacing each number in one line each one...

f=input()
ften=f.replace("10","ten")
f1=ften.replace("1","one")
f2=f1.replace("2","two")
f3=f2.replace("3","three")
f4=f3.replace("4","four")
f5=f4.replace("5","five")
f6=f5.replace("6","six")
f7=f6.replace("7","seven")
f8=f7.replace("8","eight")
f9=f8.replace("9","nine")
f0=f9.replace("0","zero")
print(f0)

but how i can do it better?

Answer 1

x = input()

dic = {
"10": "ten",
"0": "zero",
"1": "one",
"2": "two",
"3": "three",
"4": "four",
"5": "five",
"6": "six",
"7": "seven",
"8": "eight",
"9": "nine",
}

for key, value in dic.items():
x = x.replace(key, value)

print(x)

Answer 2

Thanks to say
I have problem with that

Answer 3

Funny. I did it about like that and passed anyway, although Kitten must be right.

Answer 4

How? That is the question

Answer 5

For what is required it's a good solution.
For a slightly neater way, I first defined a dictionary with all numbers in it as keys and all names of numbers in it as values.
I also defined a string '0123456789'
Then iterated through the input finding elements that are in the string of numbers and replacing them with their corresponding value in the dictionary.
Make sure to replace '10' with 'ten' before this for loop though.

Side note, I first used: for i in range(len(input)): and iterated over with input[i], but this didn't finish the job as the length of the string was longer by the end so it never reached the last digit/s
Make use of: for i in input:...

Answer 6

But 10 is onezero... or not?

Answer 7

All numbers? They are infinite :P
And what happens with the 10, 11, ...?

Answer 8

David Martínez I only did 0-10 as the question asked, I’m sure there would be a way to extend this. Maybe check: if input[i+1] in string and then change to the tens word for that number. And if you go from the end of the string backwards I think is would be easier to implement

Answer 9

David Martínez, if you make sure that 10 is replaced *first*, after that transformation only genuine 1s remain.

Answer 10

You can replace the 10 separately and then write a loop for the rest.

This was my version:

inp=input().replace('10', 'ten')
for a, b in zip(
(str(i) for i in range(10)),
'zero one two three four five six seven eight nine'.split()):
inp = inp.replace(a, b)
print(inp)

Answer 11

And the same thing for 11, 12, 13...

Answer 12

No, the task description says that we aren't supposed to touch everything beyond ten.

Answer 13

Hi Guys.
Nice debate....
Why bother with a for loop when you have a dict ready to call the values.
A string input was the request.
Dict = {'1': 'one',
'2': 'two',
Etc,
'10': 'ten', }
If the input refers directly to the key of the dictionary, you can print out the value

Answer 14

y = input().split()
nums=list('123456789')+['10']
for x in range(len(y)):
if y[x] in nums:
y[x] = y[x].replace('1','one')
y[x] = y[x].replace('2','two')
y[x] = y[x].replace('3','three')
y[x] = y[x].replace('4','four')
y[x] = y[x].replace('5','five')
y[x] = y[x].replace('6','six')
y[x] = y[x].replace('7','seven')
y[x] = y[x].replace('8','eight')
y[x] = y[x].replace('9','nine')

print(' '.join(y))

Answer 15

Guys why don't u make a translator
I made this it had one problem replacing 10 but thanks to Honfu
Suggestion it is solved check this code
import re

numwords = ["zero","one","two","three","four","five","six","seven","eight","nine","ten"]

le = input().replace("10",numwords[10])

def convertor(phrase):
translated = ""
for letter in phrase:
if letter in "0123456789" :
translated += numwords[int(letter)]
else:
translated += letter
else:
translated.replace("onezero",str(numwords[10]))
return translated

print(convertor(le))

Answer 16

My solution:

sentence = input()

import re

if re.search("10", sentence):
sentence = re.sub(r"\b10\b","ten", sentence)

if re.search("1", sentence):
sentence = re.sub(r"\b1\b","one", sentence)

if re.search("2", sentence):
sentence = re.sub(r"\b2\b","two", sentence)

if re.search("3", sentence):
sentence = re.sub(r"\b3\b","three", sentence)

if re.search("4", sentence):
sentence = re.sub(r"\b(4)\b","four", sentence)

if re.search("5", sentence):
sentence = re.sub(r"\b(5)\b","five", sentence)

if re.search("6", sentence):
sentence = re.sub(r"\b(6)\b","six", sentence)

if re.search("7", sentence):
sentence = re.sub(r"\b(7)\b","seven", sentence)

if re.search("8", sentence):
sentence = re.sub(r"\b(8)\b","eight", sentence)

if re.search("9", sentence):
sentence = re.sub(r"\b(9)\b","nine", sentence)

print(sentence)

Please tell me what are thinking about my solution.

Answer 17

I think to replace 0 to 10 we should do it with list

arr = ['zero','one','two','three','four','five','six','seven','eight','nine','ten']

st = input()
for i in range(11)[::-1]:
st = st.replace(str(i),arr[i])
print(st)

Answer 18

What's wrong with my code
I can't pass the 3rd test case
string_1 = input()
string = string_1.split()
no = "1234567890"
extract_index = []
act_no = []

for i in string:
if i in no:
act_no += i
no_index = string.index(i)
extract_index.append(no_index)

dict = {'1': "one", '2': "two", '3': "three", '4': "four", '5': "five", '6': "six", '7': "seven", '8': "eight", '9':"nine", '0': "zero",'10':"ten"}

for i in range(len(act_no)):
string[extract_index[i]] = dict.get(act_no[i])

for i in string:
print(i,end=" ")

Answer 19

phrase=input()
list=phrase.split(' ')
dict= {
'0':"zero",
'1':"one",
'2':"two",
'3':"three",
'4':"four",
'5':"five",
'6':"six",
'7':"seven",
'8':"eight",
'9':"nine",
'10':"ten"
}
output=[]
for i in list:
if i in dict:
x=dict[i]
output.append(x)
else:
output.append(i)

result=" ".join(output)
print(result)

Answer 20

I made this option

txt=list(input().replace("10","ten"))

new_txt=""

num_dict={
0:"Zero",
1:"one",
2:"two",
3:"three",
4:"four",
5:"five",
6:"six",
7:"seven",
8:"eight",
9:"nine"
}

for i in txt:
try:
new_txt+=num_dict.get(int(i))
except ValueError:
new_txt+=i

print(new_txt)

I don't know why I did int in dictionary