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?
python-3.x pandas dataframe
asked Nov 22 at 20:26
FunnyChef
6322615
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?
python-3.x pandas dataframe
asked Nov 22 at 20:26
FunnyChef
6322615
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?
python-3.x pandas dataframe
asked Nov 22 at 20:26
FunnyChef
6322615
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
python-3.x pandas dataframe
asked Nov 22 at 20:26
FunnyChef
6322615
asked Nov 22 at 20:26
FunnyChef
6322615
asked Nov 22 at 20:26
FunnyChef
6322615
asked Nov 22 at 20:26
FunnyChef
6322615
asked Nov 22 at 20:26
FunnyChef
6322615
6322615
add a comment |
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)
answered Nov 22 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 at 21:03
Julian Peller
849511
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 at 20:45
Chiheb Nexus
4,72031527
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
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)
answered Nov 22 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)
answered Nov 22 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)
answered Nov 22 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)
answered Nov 22 at 20:37
Esteban Quiros
1015
edited Nov 22 at 20:42
answered Nov 22 at 20:37
Esteban Quiros
1015
answered Nov 22 at 20:37
Esteban Quiros
1015
answered Nov 22 at 20:37
Esteban Quiros
1015
1015
add a comment |
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 at 21:03
Julian Peller
849511
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 at 21:03
Julian Peller
849511
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 at 21:03
Julian Peller
849511
With .str accessor you can achieve it in one line:
(df[0].str.startswith(" ") | df[0].str.endswith(" ")).sum()
answered Nov 22 at 21:03
Julian Peller
849511
answered Nov 22 at 21:03
Julian Peller
849511
answered Nov 22 at 21:03
Julian Peller
849511
answered Nov 22 at 21:03
Julian Peller
849511
849511
add a comment |
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 at 20:45
Chiheb Nexus
4,72031527
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 at 20:45
Chiheb Nexus
4,72031527
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 at 20:45
Chiheb Nexus
4,72031527
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 at 20:45
Chiheb Nexus
4,72031527
answered Nov 22 at 20:45
Chiheb Nexus
4,72031527
answered Nov 22 at 20:45
Chiheb Nexus
4,72031527
answered Nov 22 at 20:45
Chiheb Nexus
4,72031527
4,72031527
add a comment |
add a comment |
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