Python Count Leading and Trailing Whitespace

I have the following dataframe note the leading and trailing whitespace in the stings:

import pandas as pd

data = ['foo ', ' bar', ' baz ', 'beetle juice']

df = pd.DataFrame(data)

I need to count all strings that have leading andor trailing whitespace but ignore whitespace in the middle of the sting.

So, in the example above, the whitespace count should equal 3.

What's the best way to do this?

asked Nov 22 '18 at 20:26

FunnyChef

6402615

add a comment |

I have the following dataframe note the leading and trailing whitespace in the stings:

import pandas as pd

data = ['foo ', ' bar', ' baz ', 'beetle juice']

df = pd.DataFrame(data)

I need to count all strings that have leading andor trailing whitespace but ignore whitespace in the middle of the sting.

So, in the example above, the whitespace count should equal 3.

What's the best way to do this?

asked Nov 22 '18 at 20:26

FunnyChef

6402615

add a comment |

I have the following dataframe note the leading and trailing whitespace in the stings:

import pandas as pd

data = ['foo ', ' bar', ' baz ', 'beetle juice']

df = pd.DataFrame(data)

I need to count all strings that have leading andor trailing whitespace but ignore whitespace in the middle of the sting.

So, in the example above, the whitespace count should equal 3.

What's the best way to do this?

asked Nov 22 '18 at 20:26

FunnyChef

6402615

I have the following dataframe note the leading and trailing whitespace in the stings:

import pandas as pd

data = ['foo ', ' bar', ' baz ', 'beetle juice']

df = pd.DataFrame(data)

I need to count all strings that have leading andor trailing whitespace but ignore whitespace in the middle of the sting.

So, in the example above, the whitespace count should equal 3.

What's the best way to do this?

python-3.x pandas dataframe

asked Nov 22 '18 at 20:26

FunnyChef

6402615

asked Nov 22 '18 at 20:26

FunnyChef

6402615

asked Nov 22 '18 at 20:26

FunnyChef

6402615

asked Nov 22 '18 at 20:26

FunnyChef

6402615

asked Nov 22 '18 at 20:26

FunnyChef

6402615

add a comment |

3 Answers
3

active

oldest

votes

This code does what you want.

import pandas as pd



data = ['foo ', ' bar', ' baz ', 'beetle juice']



df = pd.DataFrame(data)

count = 0



for i,row in df.iterrows():

    if row[0][0] == " " or row[0][-1] == " ":

        count += 1



print(count)

edited Nov 22 '18 at 20:42

answered Nov 22 '18 at 20:37

Esteban Quiros

1015

add a comment |

With .str accessor you can achieve it in one line:

(df[0].str.startswith(" ") | df[0].str.endswith(" ")).sum()

answered Nov 22 '18 at 21:03

Julian Peller

8941511

add a comment |

Here is a solution using defaultdict from collection module:

from collections import defaultdict as df



data = ['foo ', ' bar', ' baz ', 'beetle juice']

result = df(int)



for elm in data:

    if elm.startswith(' '):

        result['leading'] += 1

    elif elm.endswith(' '):

        result['trailing'] += 1



print(result)

print(dict(result))

count = sum(k for k in result.values())

print(count)

Output:

defaultdict(<class 'int'>, {'trailing': 1, 'leading': 2})

{'trailing': 1, 'leading': 2}

3

answered Nov 22 '18 at 20:45

Chiheb Nexus

5,01031627

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

This code does what you want.

import pandas as pd



data = ['foo ', ' bar', ' baz ', 'beetle juice']



df = pd.DataFrame(data)

count = 0



for i,row in df.iterrows():

    if row[0][0] == " " or row[0][-1] == " ":

        count += 1



print(count)

edited Nov 22 '18 at 20:42

answered Nov 22 '18 at 20:37

Esteban Quiros

1015

add a comment |

This code does what you want.

import pandas as pd



data = ['foo ', ' bar', ' baz ', 'beetle juice']



df = pd.DataFrame(data)

count = 0



for i,row in df.iterrows():

    if row[0][0] == " " or row[0][-1] == " ":

        count += 1



print(count)

edited Nov 22 '18 at 20:42

answered Nov 22 '18 at 20:37

Esteban Quiros

1015

add a comment |

This code does what you want.

import pandas as pd



data = ['foo ', ' bar', ' baz ', 'beetle juice']



df = pd.DataFrame(data)

count = 0



for i,row in df.iterrows():

    if row[0][0] == " " or row[0][-1] == " ":

        count += 1



print(count)

edited Nov 22 '18 at 20:42

answered Nov 22 '18 at 20:37

Esteban Quiros

1015

This code does what you want.

import pandas as pd



data = ['foo ', ' bar', ' baz ', 'beetle juice']



df = pd.DataFrame(data)

count = 0



for i,row in df.iterrows():

    if row[0][0] == " " or row[0][-1] == " ":

        count += 1



print(count)

edited Nov 22 '18 at 20:42

answered Nov 22 '18 at 20:37

Esteban Quiros

1015

edited Nov 22 '18 at 20:42

answered Nov 22 '18 at 20:37

Esteban Quiros

1015

answered Nov 22 '18 at 20:37

Esteban Quiros

1015

answered Nov 22 '18 at 20:37

Esteban Quiros

1015

add a comment |

With .str accessor you can achieve it in one line:

(df[0].str.startswith(" ") | df[0].str.endswith(" ")).sum()

answered Nov 22 '18 at 21:03

Julian Peller

8941511

add a comment |

With .str accessor you can achieve it in one line:

(df[0].str.startswith(" ") | df[0].str.endswith(" ")).sum()

answered Nov 22 '18 at 21:03

Julian Peller

8941511

add a comment |

With .str accessor you can achieve it in one line:

(df[0].str.startswith(" ") | df[0].str.endswith(" ")).sum()

answered Nov 22 '18 at 21:03

Julian Peller

8941511

With .str accessor you can achieve it in one line:

(df[0].str.startswith(" ") | df[0].str.endswith(" ")).sum()

answered Nov 22 '18 at 21:03

Julian Peller

8941511

answered Nov 22 '18 at 21:03

Julian Peller

8941511

answered Nov 22 '18 at 21:03

Julian Peller

8941511

answered Nov 22 '18 at 21:03

Julian Peller

8941511

add a comment |

Here is a solution using defaultdict from collection module:

from collections import defaultdict as df



data = ['foo ', ' bar', ' baz ', 'beetle juice']

result = df(int)



for elm in data:

    if elm.startswith(' '):

        result['leading'] += 1

    elif elm.endswith(' '):

        result['trailing'] += 1



print(result)

print(dict(result))

count = sum(k for k in result.values())

print(count)

Output:

defaultdict(<class 'int'>, {'trailing': 1, 'leading': 2})

{'trailing': 1, 'leading': 2}

3

answered Nov 22 '18 at 20:45

Chiheb Nexus

5,01031627

add a comment |

Here is a solution using defaultdict from collection module:

from collections import defaultdict as df



data = ['foo ', ' bar', ' baz ', 'beetle juice']

result = df(int)



for elm in data:

    if elm.startswith(' '):

        result['leading'] += 1

    elif elm.endswith(' '):

        result['trailing'] += 1



print(result)

print(dict(result))

count = sum(k for k in result.values())

print(count)

Output:

defaultdict(<class 'int'>, {'trailing': 1, 'leading': 2})

{'trailing': 1, 'leading': 2}

3

answered Nov 22 '18 at 20:45

Chiheb Nexus

5,01031627

add a comment |

Here is a solution using defaultdict from collection module:

from collections import defaultdict as df



data = ['foo ', ' bar', ' baz ', 'beetle juice']

result = df(int)



for elm in data:

    if elm.startswith(' '):

        result['leading'] += 1

    elif elm.endswith(' '):

        result['trailing'] += 1



print(result)

print(dict(result))

count = sum(k for k in result.values())

print(count)

Output:

defaultdict(<class 'int'>, {'trailing': 1, 'leading': 2})

{'trailing': 1, 'leading': 2}

3

answered Nov 22 '18 at 20:45

Chiheb Nexus

5,01031627

Here is a solution using defaultdict from collection module:

from collections import defaultdict as df



data = ['foo ', ' bar', ' baz ', 'beetle juice']

result = df(int)



for elm in data:

    if elm.startswith(' '):

        result['leading'] += 1

    elif elm.endswith(' '):

        result['trailing'] += 1



print(result)

print(dict(result))

count = sum(k for k in result.values())

print(count)

Output:

defaultdict(<class 'int'>, {'trailing': 1, 'leading': 2})

{'trailing': 1, 'leading': 2}

3

answered Nov 22 '18 at 20:45

Chiheb Nexus

5,01031627

answered Nov 22 '18 at 20:45

Chiheb Nexus

5,01031627

answered Nov 22 '18 at 20:45

Chiheb Nexus

5,01031627

answered Nov 22 '18 at 20:45

Chiheb Nexus

5,01031627

add a comment |

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