check if a string contains any char from set

-1

Actually, i'm working on a task from SPOJ. How to check if a string contains any char from set, but first char from string where a char from set occur can not be deleted.

F.e.
Have a string

word = "anAconda_elEphant"

and a set of vowels:

vowels = set('aeiouyAEIOUY')

I want in result a string

word = "ancnd_lphnt"

This should return True when occurence of any char in set is equal 1. I know that argument for a method .count() must be str, not set.

if word.count(vowels) == 1:

   for char in word[char_pos:]:

        if char in vowels:

            char.replace('')

edited Nov 21 '18 at 23:29

ewwink

11.6k22238

asked Nov 21 '18 at 23:25

Radosław Hryniewicki

Do you want to return a True/False result, or return the modified string?

– Barmar
Nov 21 '18 at 23:33

1

When you say the first char shouldn't be deleted, do you mean the first vowel shouldn't be deleted, or the first char of any type? E.g. what should be the result for word = "bace", bc or bac?

– Barmar
Nov 21 '18 at 23:35

if its bac that makes my solution invalid

– Joran Beasley
Nov 21 '18 at 23:36

@Barmar want a return an int which tell me how many vowels are in string. True = 1, False !=1

– Radosław Hryniewicki
Nov 22 '18 at 8:32

But you also said you want to delete vowels from the string. Do you want two returns, the count of vowels and the new string?

– Barmar
Nov 23 '18 at 6:18

add a comment |

-1

Actually, i'm working on a task from SPOJ. How to check if a string contains any char from set, but first char from string where a char from set occur can not be deleted.

F.e.
Have a string

word = "anAconda_elEphant"

and a set of vowels:

vowels = set('aeiouyAEIOUY')

I want in result a string

word = "ancnd_lphnt"

This should return True when occurence of any char in set is equal 1. I know that argument for a method .count() must be str, not set.

if word.count(vowels) == 1:

   for char in word[char_pos:]:

        if char in vowels:

            char.replace('')

edited Nov 21 '18 at 23:29

ewwink

11.6k22238

asked Nov 21 '18 at 23:25

Radosław Hryniewicki

Do you want to return a True/False result, or return the modified string?

– Barmar
Nov 21 '18 at 23:33

1

When you say the first char shouldn't be deleted, do you mean the first vowel shouldn't be deleted, or the first char of any type? E.g. what should be the result for word = "bace", bc or bac?

– Barmar
Nov 21 '18 at 23:35

if its bac that makes my solution invalid

– Joran Beasley
Nov 21 '18 at 23:36

@Barmar want a return an int which tell me how many vowels are in string. True = 1, False !=1

– Radosław Hryniewicki
Nov 22 '18 at 8:32

But you also said you want to delete vowels from the string. Do you want two returns, the count of vowels and the new string?

– Barmar
Nov 23 '18 at 6:18

add a comment |

-1

Actually, i'm working on a task from SPOJ. How to check if a string contains any char from set, but first char from string where a char from set occur can not be deleted.

F.e.
Have a string

word = "anAconda_elEphant"

and a set of vowels:

vowels = set('aeiouyAEIOUY')

I want in result a string

word = "ancnd_lphnt"

This should return True when occurence of any char in set is equal 1. I know that argument for a method .count() must be str, not set.

if word.count(vowels) == 1:

   for char in word[char_pos:]:

        if char in vowels:

            char.replace('')

edited Nov 21 '18 at 23:29

ewwink

11.6k22238

asked Nov 21 '18 at 23:25

Radosław Hryniewicki

Actually, i'm working on a task from SPOJ. How to check if a string contains any char from set, but first char from string where a char from set occur can not be deleted.

F.e.
Have a string

word = "anAconda_elEphant"

and a set of vowels:

vowels = set('aeiouyAEIOUY')

I want in result a string

word = "ancnd_lphnt"

This should return True when occurence of any char in set is equal 1. I know that argument for a method .count() must be str, not set.

if word.count(vowels) == 1:

   for char in word[char_pos:]:

        if char in vowels:

            char.replace('')

python

edited Nov 21 '18 at 23:29

ewwink

11.6k22238

asked Nov 21 '18 at 23:25

Radosław Hryniewicki

edited Nov 21 '18 at 23:29

ewwink

11.6k22238

asked Nov 21 '18 at 23:25

Radosław Hryniewicki

edited Nov 21 '18 at 23:29

ewwink

11.6k22238

edited Nov 21 '18 at 23:29

ewwink

11.6k22238

edited Nov 21 '18 at 23:29

ewwink

11.6k22238

asked Nov 21 '18 at 23:25

Radosław Hryniewicki

asked Nov 21 '18 at 23:25

Radosław Hryniewicki

asked Nov 21 '18 at 23:25

Radosław Hryniewicki

Do you want to return a True/False result, or return the modified string?

– Barmar
Nov 21 '18 at 23:33

1

When you say the first char shouldn't be deleted, do you mean the first vowel shouldn't be deleted, or the first char of any type? E.g. what should be the result for word = "bace", bc or bac?

– Barmar
Nov 21 '18 at 23:35

if its bac that makes my solution invalid

– Joran Beasley
Nov 21 '18 at 23:36

@Barmar want a return an int which tell me how many vowels are in string. True = 1, False !=1

– Radosław Hryniewicki
Nov 22 '18 at 8:32

But you also said you want to delete vowels from the string. Do you want two returns, the count of vowels and the new string?

– Barmar
Nov 23 '18 at 6:18

add a comment |

Do you want to return a True/False result, or return the modified string?

– Barmar
Nov 21 '18 at 23:33

1

When you say the first char shouldn't be deleted, do you mean the first vowel shouldn't be deleted, or the first char of any type? E.g. what should be the result for word = "bace", bc or bac?

– Barmar
Nov 21 '18 at 23:35

if its bac that makes my solution invalid

– Joran Beasley
Nov 21 '18 at 23:36

@Barmar want a return an int which tell me how many vowels are in string. True = 1, False !=1

– Radosław Hryniewicki
Nov 22 '18 at 8:32

But you also said you want to delete vowels from the string. Do you want two returns, the count of vowels and the new string?

– Barmar
Nov 23 '18 at 6:18

Do you want to return a True/False result, or return the modified string?

– Barmar
Nov 21 '18 at 23:33

When you say the first char shouldn't be deleted, do you mean the first vowel shouldn't be deleted, or the first char of any type? E.g. what should be the result for word = "bace", bc or bac?

– Barmar
Nov 21 '18 at 23:35

if its bac that makes my solution invalid

– Joran Beasley
Nov 21 '18 at 23:36

@Barmar want a return an int which tell me how many vowels are in string. True = 1, False !=1

– Radosław Hryniewicki
Nov 22 '18 at 8:32

But you also said you want to delete vowels from the string. Do you want two returns, the count of vowels and the new string?

– Barmar
Nov 23 '18 at 6:18

add a comment |

3 Answers
3

active

oldest

votes

just use a regular expression

import re

word = "anAconda_elEphant"

# use a  "lookbehind" to make sure there is at least one character in front of this character...

print(re.sub("(?<=.)[aeiouyAEIOUY]",'',word))

# 'ancnd_lphnt'

as mentioned if you expect it to skip the first match of the set as opposed to just the first letter you will need a different solution

print(re.sub("(?<=.)[aeiouyAEIOUY]",'',"bace"))

# 'bc' # a is not the FIRST letter so it is replaced

the easiest is to split it into two steps
first split the string on the first match

word = "bace"

splitted_string = re.split("(.*?[aeiouyAEIOUY])",word,1)

# you will notice we have an extra empty string at the beginning of our matches ... so we can skip that

lhs,rhs = splitted_string[1:]

# now just run a simple re.sub on our rhs and rejoin the halves

print(lhs + re.sub("[aeiouyAEIOUY]",'',rhs))

# results in "bac"

edited Nov 21 '18 at 23:44

answered Nov 21 '18 at 23:33

Joran Beasley

72.7k678118

Wished I had the mental capacity to finangle with my own REGEX. Nice answer

– Jaba
Nov 21 '18 at 23:42

add a comment |

You can use a for loop as below. The idea is to build a list, and use a flag to mark when you meet a character from vowels.

word = "anAconda_elEphant"

vowels = set('aeiouyAEIOUY')



flag = False



L = 

for ch in word:

    if (ch not in vowels) or (not flag):

        L.append(ch)

    if ch in vowels:

        flag = True



word = ''.join(L)



print(word)



ancnd_lphnt

edited Nov 21 '18 at 23:41

answered Nov 21 '18 at 23:31

jpp

96.7k2158109

add a comment |

print(vowels.intersection(word))

https://docs.python.org/3/library/stdtypes.html#frozenset.intersection

answered Nov 22 '18 at 0:28

rikAtee

4,83552958

He wants to remove the vowels, so it should be word.difference(vowels)

– Barmar
Nov 23 '18 at 6:20

actually, neither would work if there are duplicate consonants - only one of the duplicates would survive.

– rikAtee
Nov 23 '18 at 14:43

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

just use a regular expression

import re

word = "anAconda_elEphant"

# use a  "lookbehind" to make sure there is at least one character in front of this character...

print(re.sub("(?<=.)[aeiouyAEIOUY]",'',word))

# 'ancnd_lphnt'

as mentioned if you expect it to skip the first match of the set as opposed to just the first letter you will need a different solution

print(re.sub("(?<=.)[aeiouyAEIOUY]",'',"bace"))

# 'bc' # a is not the FIRST letter so it is replaced

the easiest is to split it into two steps
first split the string on the first match

word = "bace"

splitted_string = re.split("(.*?[aeiouyAEIOUY])",word,1)

# you will notice we have an extra empty string at the beginning of our matches ... so we can skip that

lhs,rhs = splitted_string[1:]

# now just run a simple re.sub on our rhs and rejoin the halves

print(lhs + re.sub("[aeiouyAEIOUY]",'',rhs))

# results in "bac"

edited Nov 21 '18 at 23:44

answered Nov 21 '18 at 23:33

Joran Beasley

72.7k678118

Wished I had the mental capacity to finangle with my own REGEX. Nice answer

– Jaba
Nov 21 '18 at 23:42

add a comment |

just use a regular expression

import re

word = "anAconda_elEphant"

# use a  "lookbehind" to make sure there is at least one character in front of this character...

print(re.sub("(?<=.)[aeiouyAEIOUY]",'',word))

# 'ancnd_lphnt'

as mentioned if you expect it to skip the first match of the set as opposed to just the first letter you will need a different solution

print(re.sub("(?<=.)[aeiouyAEIOUY]",'',"bace"))

# 'bc' # a is not the FIRST letter so it is replaced

the easiest is to split it into two steps
first split the string on the first match

word = "bace"

splitted_string = re.split("(.*?[aeiouyAEIOUY])",word,1)

# you will notice we have an extra empty string at the beginning of our matches ... so we can skip that

lhs,rhs = splitted_string[1:]

# now just run a simple re.sub on our rhs and rejoin the halves

print(lhs + re.sub("[aeiouyAEIOUY]",'',rhs))

# results in "bac"

edited Nov 21 '18 at 23:44

answered Nov 21 '18 at 23:33

Joran Beasley

72.7k678118

Wished I had the mental capacity to finangle with my own REGEX. Nice answer

– Jaba
Nov 21 '18 at 23:42

add a comment |

just use a regular expression

import re

word = "anAconda_elEphant"

# use a  "lookbehind" to make sure there is at least one character in front of this character...

print(re.sub("(?<=.)[aeiouyAEIOUY]",'',word))

# 'ancnd_lphnt'

as mentioned if you expect it to skip the first match of the set as opposed to just the first letter you will need a different solution

print(re.sub("(?<=.)[aeiouyAEIOUY]",'',"bace"))

# 'bc' # a is not the FIRST letter so it is replaced

the easiest is to split it into two steps
first split the string on the first match

word = "bace"

splitted_string = re.split("(.*?[aeiouyAEIOUY])",word,1)

# you will notice we have an extra empty string at the beginning of our matches ... so we can skip that

lhs,rhs = splitted_string[1:]

# now just run a simple re.sub on our rhs and rejoin the halves

print(lhs + re.sub("[aeiouyAEIOUY]",'',rhs))

# results in "bac"

edited Nov 21 '18 at 23:44

answered Nov 21 '18 at 23:33

Joran Beasley

72.7k678118

just use a regular expression

import re

word = "anAconda_elEphant"

# use a  "lookbehind" to make sure there is at least one character in front of this character...

print(re.sub("(?<=.)[aeiouyAEIOUY]",'',word))

# 'ancnd_lphnt'

as mentioned if you expect it to skip the first match of the set as opposed to just the first letter you will need a different solution

print(re.sub("(?<=.)[aeiouyAEIOUY]",'',"bace"))

# 'bc' # a is not the FIRST letter so it is replaced

the easiest is to split it into two steps
first split the string on the first match

word = "bace"

splitted_string = re.split("(.*?[aeiouyAEIOUY])",word,1)

# you will notice we have an extra empty string at the beginning of our matches ... so we can skip that

lhs,rhs = splitted_string[1:]

# now just run a simple re.sub on our rhs and rejoin the halves

print(lhs + re.sub("[aeiouyAEIOUY]",'',rhs))

# results in "bac"

edited Nov 21 '18 at 23:44

answered Nov 21 '18 at 23:33

Joran Beasley

72.7k678118

edited Nov 21 '18 at 23:44

answered Nov 21 '18 at 23:33

Joran Beasley

72.7k678118

answered Nov 21 '18 at 23:33

Joran Beasley

72.7k678118

answered Nov 21 '18 at 23:33

Joran Beasley

72.7k678118

Wished I had the mental capacity to finangle with my own REGEX. Nice answer

– Jaba
Nov 21 '18 at 23:42

add a comment |

Wished I had the mental capacity to finangle with my own REGEX. Nice answer

– Jaba
Nov 21 '18 at 23:42

Wished I had the mental capacity to finangle with my own REGEX. Nice answer

– Jaba
Nov 21 '18 at 23:42

add a comment |

You can use a for loop as below. The idea is to build a list, and use a flag to mark when you meet a character from vowels.

word = "anAconda_elEphant"

vowels = set('aeiouyAEIOUY')



flag = False



L = 

for ch in word:

    if (ch not in vowels) or (not flag):

        L.append(ch)

    if ch in vowels:

        flag = True



word = ''.join(L)



print(word)



ancnd_lphnt

edited Nov 21 '18 at 23:41

answered Nov 21 '18 at 23:31

jpp

96.7k2158109

add a comment |

You can use a for loop as below. The idea is to build a list, and use a flag to mark when you meet a character from vowels.

word = "anAconda_elEphant"

vowels = set('aeiouyAEIOUY')



flag = False



L = 

for ch in word:

    if (ch not in vowels) or (not flag):

        L.append(ch)

    if ch in vowels:

        flag = True



word = ''.join(L)



print(word)



ancnd_lphnt

edited Nov 21 '18 at 23:41

answered Nov 21 '18 at 23:31

jpp

96.7k2158109

add a comment |

You can use a for loop as below. The idea is to build a list, and use a flag to mark when you meet a character from vowels.

word = "anAconda_elEphant"

vowels = set('aeiouyAEIOUY')



flag = False



L = 

for ch in word:

    if (ch not in vowels) or (not flag):

        L.append(ch)

    if ch in vowels:

        flag = True



word = ''.join(L)



print(word)



ancnd_lphnt

edited Nov 21 '18 at 23:41

answered Nov 21 '18 at 23:31

jpp

96.7k2158109

You can use a for loop as below. The idea is to build a list, and use a flag to mark when you meet a character from vowels.

word = "anAconda_elEphant"

vowels = set('aeiouyAEIOUY')



flag = False



L = 

for ch in word:

    if (ch not in vowels) or (not flag):

        L.append(ch)

    if ch in vowels:

        flag = True



word = ''.join(L)



print(word)



ancnd_lphnt

edited Nov 21 '18 at 23:41

answered Nov 21 '18 at 23:31

jpp

96.7k2158109

edited Nov 21 '18 at 23:41

answered Nov 21 '18 at 23:31

jpp

96.7k2158109

answered Nov 21 '18 at 23:31

jpp

96.7k2158109

answered Nov 21 '18 at 23:31

jpp

96.7k2158109

add a comment |

print(vowels.intersection(word))

https://docs.python.org/3/library/stdtypes.html#frozenset.intersection

answered Nov 22 '18 at 0:28

rikAtee

4,83552958

He wants to remove the vowels, so it should be word.difference(vowels)

– Barmar
Nov 23 '18 at 6:20

actually, neither would work if there are duplicate consonants - only one of the duplicates would survive.

– rikAtee
Nov 23 '18 at 14:43

add a comment |

print(vowels.intersection(word))

https://docs.python.org/3/library/stdtypes.html#frozenset.intersection

answered Nov 22 '18 at 0:28

rikAtee

4,83552958

He wants to remove the vowels, so it should be word.difference(vowels)

– Barmar
Nov 23 '18 at 6:20

actually, neither would work if there are duplicate consonants - only one of the duplicates would survive.

– rikAtee
Nov 23 '18 at 14:43

add a comment |

print(vowels.intersection(word))

https://docs.python.org/3/library/stdtypes.html#frozenset.intersection

answered Nov 22 '18 at 0:28

rikAtee

4,83552958

print(vowels.intersection(word))

https://docs.python.org/3/library/stdtypes.html#frozenset.intersection

answered Nov 22 '18 at 0:28

rikAtee

4,83552958

answered Nov 22 '18 at 0:28

rikAtee

4,83552958

answered Nov 22 '18 at 0:28

rikAtee

4,83552958

answered Nov 22 '18 at 0:28

rikAtee

4,83552958

He wants to remove the vowels, so it should be word.difference(vowels)

– Barmar
Nov 23 '18 at 6:20

actually, neither would work if there are duplicate consonants - only one of the duplicates would survive.

– rikAtee
Nov 23 '18 at 14:43

add a comment |

He wants to remove the vowels, so it should be word.difference(vowels)

– Barmar
Nov 23 '18 at 6:20

actually, neither would work if there are duplicate consonants - only one of the duplicates would survive.

– rikAtee
Nov 23 '18 at 14:43

He wants to remove the vowels, so it should be word.difference(vowels)

– Barmar
Nov 23 '18 at 6:20

actually, neither would work if there are duplicate consonants - only one of the duplicates would survive.

– rikAtee
Nov 23 '18 at 14:43

add a comment |

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