Easiest way to calculate number of character in string

[P. Schmidt-Volkmar]

Hi there,

I have a string in which I want to calculate how often the character ';'
occurs. If the character does not occur 42 times, the ";" should be added so
the 42 are reached.

My solution is slow and wrong:
for Position in range (0, len(Zeile)):
if Zeile[Position]==';': AnzahlSemikolon = AnzahlSemikolon +1
if AnzahlSemikolon < 42:
for Zaehler in range(AnzahlSemikolon, 42):
Zeile = Zeile + ';'
Dreckskram = Dreckskram +1

How can this be achieved easily?

Thanks,

Pascal

 

[claude.henchoz]

should = 42
has = Zeile.count(';')
if has < should:
Zeile += ";"*(should - has)

cheers, claude

 

[Carsten Haese]

Hi there,

I have a string in which I want to calculate how often the character ';'
occurs. If the character does not occur 42 times, the ";" should be added so
the 42 are reached.

My solution is slow and wrong:
for Position in range (0, len(Zeile)):
if Zeile[Position]==';': AnzahlSemikolon = AnzahlSemikolon +1
if AnzahlSemikolon < 42:
for Zaehler in range(AnzahlSemikolon, 42):
Zeile = Zeile + ';'
Dreckskram = Dreckskram +1

How can this be achieved easily?

Thanks,

Pascal

AnzahlSemikolon = Zeile.count(';')
if AnzahlSemikolon < 42:
Zeile += ';'*(42-AnzahlSemikolon)

HTH,

Carsten.

 

[Lawrence Oluyede]

Hi there,

I have a string in which I want to calculate how often the character ';'
occurs. If the character does not occur 42 times, the ";" should be added so
the 42 are reached.

My solution is slow and wrong:
for Position in range (0, len(Zeile)):
if Zeile[Position]==';': AnzahlSemikolon = AnzahlSemikolon +1
if AnzahlSemikolon < 42:
for Zaehler in range(AnzahlSemikolon, 42):
Zeile = Zeile + ';'
Dreckskram = Dreckskram +1

How can this be achieved easily?

One way can be this I think:

s = "your_string_full_of_;;;;;"
num_of_semicolons = len(s.split(";") - 1)

 

[Kent Johnson]

Hi there,

I have a string in which I want to calculate how often the character ';'
occurs. If the character does not occur 42 times, the ";" should be added so
the 42 are reached.

My solution is slow and wrong:

How can this be achieved easily?

Is this homework? str.count() and string multiplication are your friends here. See
http://docs.python.org/lib/string-methods.html
http://docs.python.org/lib/typesseq.html

Kent

 

[Ove Svensson]

Hi there,

I have a string in which I want to calculate how often the character ';'
occurs. If the character does not occur 42 times, the ";" should be added so
the 42 are reached.

My solution is slow and wrong:
for Position in range (0, len(Zeile)):
if Zeile[Position]==';': AnzahlSemikolon = AnzahlSemikolon +1
if AnzahlSemikolon < 42:
for Zaehler in range(AnzahlSemikolon, 42):
Zeile = Zeile + ';'
Dreckskram = Dreckskram +1

How can this be achieved easily?

Thanks,

Pascal

What about this:

Zaehler += ';'*max(0,42-Zaehler.count(';'))

 

[Ove Svensson]

Hi there,

I have a string in which I want to calculate how often the character ';'
occurs. If the character does not occur 42 times, the ";" should be added so
the 42 are reached.

My solution is slow and wrong:
for Position in range (0, len(Zeile)):
if Zeile[Position]==';': AnzahlSemikolon = AnzahlSemikolon +1
if AnzahlSemikolon < 42:
for Zaehler in range(AnzahlSemikolon, 42):
Zeile = Zeile + ';'
Dreckskram = Dreckskram +1

How can this be achieved easily?

Thanks,

Pascal

What about this:

Zaehler += ';'*max(0,42-Zaehler.count(';'))

Sorry, should have been

Zeile += ';'*max(0,42-Zeile.count(';'))

 

[Bengt Richter]

Hi there,

I have a string in which I want to calculate how often the character ';'
occurs. If the character does not occur 42 times, the ";" should be added so
the 42 are reached.

My solution is slow and wrong:
for Position in range (0, len(Zeile)):
if Zeile[Position]==';': AnzahlSemikolon = AnzahlSemikolon +1
if AnzahlSemikolon < 42:
for Zaehler in range(AnzahlSemikolon, 42):
Zeile = Zeile + ';'
Dreckskram = Dreckskram +1

How can this be achieved easily?

Thanks,

Pascal

What about this:

Zaehler += ';'*max(0,42-Zaehler.count(';'))

Sorry, should have been

Zeile += ';'*max(0,42-Zeile.count(';'))

I think You don't need the max
...
-3: ''
-2: ''
-1: ''
0: ''
1: ';'
2: ';;'
3: ';;;'

I.e.,

Zeile += ';'*(42-Zeile.count(';'))

should work, since a string is a sequence type and

http://docs.python.org/lib/typesseq.html

Says
"""
Operation Result Notes
...
s * n , n * s n shallow copies of s concatenated (2)
....
(2)
Values of n less than 0 are treated as 0 (which yields an empty sequence of the same type as s). ...
"""

I guess it might be nice to mention this in help(str) also, to publish a useful fact better.
Maybe under str.__mul__ ?

Regards,
Bengt Richter