How to do group by in two levels on python pandas an count values?












1














I have the following dataframe



import pandas as pd

import numpy as np

df = pd.DataFrame()

df['Name'] = ['AK', 'Ram', 'Ram', 'Singh', 'Murugan', 'Kishore', 'AK']

df['Email'] = ['AK@gmail.com', 'a@djgbj.com', 'a@djgbj.com', '3454@ghhg.io', 'dgg@qw.cc', 'dgdg@dg.com', 'AK@gmail.com']

df['Cat'] = ['ab1', 'ab2', 'ab1', 'ab2', 'ab1', 'ab2', 'ab1']

df['Id'] = ['abc1', 'abc2', 'abc3', 'abc4', 'abc5', 'abc6', 'abc7']


For the following code



dfs=df.groupby(['Email', 'Cat'])['Email'].count().reset_index(name='Number')


It gives: 



      Email         Cat Number

0   3454@ghhg.io    ab2 1

1   AK@gmail.com    ab1 2

2   a@djgbj.com     ab1 1

3   a@djgbj.com     ab2 1

4   dgdg@dg.com     ab2 1

5   dgg@qw.cc       ab1 1


How to group by on dfs to get the following output?



Cat Number Count

ab1 1      3

ab1 2      1

ab2 1      3





python pandas







share|improve this question
























  • Multiple solutions, e.g. see Pandas groupby.size vs series.value_counts vs collections.Counter with multiple series
    – jpp
    Nov 12 '18 at 11:12










  • Its should be simple with groupby and count.
    – pygo
    Nov 12 '18 at 11:32










  • I think you want c = df.groupby(['Email', 'Cat'])['Email'].cumcount(); df.groupby(['Cat', c]).Cat.count() if not then your question has been phrased poorly.
    – coldspeed
    Nov 12 '18 at 17:07
















1














I have the following dataframe



import pandas as pd

import numpy as np

df = pd.DataFrame()

df['Name'] = ['AK', 'Ram', 'Ram', 'Singh', 'Murugan', 'Kishore', 'AK']

df['Email'] = ['AK@gmail.com', 'a@djgbj.com', 'a@djgbj.com', '3454@ghhg.io', 'dgg@qw.cc', 'dgdg@dg.com', 'AK@gmail.com']

df['Cat'] = ['ab1', 'ab2', 'ab1', 'ab2', 'ab1', 'ab2', 'ab1']

df['Id'] = ['abc1', 'abc2', 'abc3', 'abc4', 'abc5', 'abc6', 'abc7']


For the following code



dfs=df.groupby(['Email', 'Cat'])['Email'].count().reset_index(name='Number')


It gives: 



      Email         Cat Number

0   3454@ghhg.io    ab2 1

1   AK@gmail.com    ab1 2

2   a@djgbj.com     ab1 1

3   a@djgbj.com     ab2 1

4   dgdg@dg.com     ab2 1

5   dgg@qw.cc       ab1 1


How to group by on dfs to get the following output?



Cat Number Count

ab1 1      3

ab1 2      1

ab2 1      3





python pandas







share|improve this question
























  • Multiple solutions, e.g. see Pandas groupby.size vs series.value_counts vs collections.Counter with multiple series
    – jpp
    Nov 12 '18 at 11:12










  • Its should be simple with groupby and count.
    – pygo
    Nov 12 '18 at 11:32










  • I think you want c = df.groupby(['Email', 'Cat'])['Email'].cumcount(); df.groupby(['Cat', c]).Cat.count() if not then your question has been phrased poorly.
    – coldspeed
    Nov 12 '18 at 17:07














1












1








1







I have the following dataframe



import pandas as pd

import numpy as np

df = pd.DataFrame()

df['Name'] = ['AK', 'Ram', 'Ram', 'Singh', 'Murugan', 'Kishore', 'AK']

df['Email'] = ['AK@gmail.com', 'a@djgbj.com', 'a@djgbj.com', '3454@ghhg.io', 'dgg@qw.cc', 'dgdg@dg.com', 'AK@gmail.com']

df['Cat'] = ['ab1', 'ab2', 'ab1', 'ab2', 'ab1', 'ab2', 'ab1']

df['Id'] = ['abc1', 'abc2', 'abc3', 'abc4', 'abc5', 'abc6', 'abc7']


For the following code



dfs=df.groupby(['Email', 'Cat'])['Email'].count().reset_index(name='Number')


It gives: 



      Email         Cat Number

0   3454@ghhg.io    ab2 1

1   AK@gmail.com    ab1 2

2   a@djgbj.com     ab1 1

3   a@djgbj.com     ab2 1

4   dgdg@dg.com     ab2 1

5   dgg@qw.cc       ab1 1


How to group by on dfs to get the following output?



Cat Number Count

ab1 1      3

ab1 2      1

ab2 1      3





python pandas







share|improve this question















I have the following dataframe



import pandas as pd

import numpy as np

df = pd.DataFrame()

df['Name'] = ['AK', 'Ram', 'Ram', 'Singh', 'Murugan', 'Kishore', 'AK']

df['Email'] = ['AK@gmail.com', 'a@djgbj.com', 'a@djgbj.com', '3454@ghhg.io', 'dgg@qw.cc', 'dgdg@dg.com', 'AK@gmail.com']

df['Cat'] = ['ab1', 'ab2', 'ab1', 'ab2', 'ab1', 'ab2', 'ab1']

df['Id'] = ['abc1', 'abc2', 'abc3', 'abc4', 'abc5', 'abc6', 'abc7']


For the following code



dfs=df.groupby(['Email', 'Cat'])['Email'].count().reset_index(name='Number')


It gives: 



      Email         Cat Number

0   3454@ghhg.io    ab2 1

1   AK@gmail.com    ab1 2

2   a@djgbj.com     ab1 1

3   a@djgbj.com     ab2 1

4   dgdg@dg.com     ab2 1

5   dgg@qw.cc       ab1 1


How to group by on dfs to get the following output?



Cat Number Count

ab1 1      3

ab1 2      1

ab2 1      3





python pandas




python pandas






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 '18 at 12:45









Ali AzG

581515




581515










asked Nov 12 '18 at 11:08









pandapanda

967




967












  • Multiple solutions, e.g. see Pandas groupby.size vs series.value_counts vs collections.Counter with multiple series
    – jpp
    Nov 12 '18 at 11:12










  • Its should be simple with groupby and count.
    – pygo
    Nov 12 '18 at 11:32










  • I think you want c = df.groupby(['Email', 'Cat'])['Email'].cumcount(); df.groupby(['Cat', c]).Cat.count() if not then your question has been phrased poorly.
    – coldspeed
    Nov 12 '18 at 17:07


















  • Multiple solutions, e.g. see Pandas groupby.size vs series.value_counts vs collections.Counter with multiple series
    – jpp
    Nov 12 '18 at 11:12










  • Its should be simple with groupby and count.
    – pygo
    Nov 12 '18 at 11:32










  • I think you want c = df.groupby(['Email', 'Cat'])['Email'].cumcount(); df.groupby(['Cat', c]).Cat.count() if not then your question has been phrased poorly.
    – coldspeed
    Nov 12 '18 at 17:07
















Multiple solutions, e.g. see Pandas groupby.size vs series.value_counts vs collections.Counter with multiple series
– jpp
Nov 12 '18 at 11:12




Multiple solutions, e.g. see Pandas groupby.size vs series.value_counts vs collections.Counter with multiple series
– jpp
Nov 12 '18 at 11:12












Its should be simple with groupby and count.
– pygo
Nov 12 '18 at 11:32




Its should be simple with groupby and count.
– pygo
Nov 12 '18 at 11:32












I think you want c = df.groupby(['Email', 'Cat'])['Email'].cumcount(); df.groupby(['Cat', c]).Cat.count() if not then your question has been phrased poorly.
– coldspeed
Nov 12 '18 at 17:07




I think you want c = df.groupby(['Email', 'Cat'])['Email'].cumcount(); df.groupby(['Cat', c]).Cat.count() if not then your question has been phrased poorly.
– coldspeed
Nov 12 '18 at 17:07












2 Answers
2






active

oldest

votes


















4














Use groupby+size and reset_index:



df1 = dfs.groupby(['Cat','Number']).size().reset_index(name='Count')


Or:



df1 = dfs.groupby(['Cat','Number'])['Email'].value_counts().reset_index(name='Count')





print(df1)

   Cat  Number  Count

0  ab1       1      2

1  ab1       2      1

2  ab2       1      3





share|improve this answer























  • Isn't your answer incorrect?
    – coldspeed
    Nov 12 '18 at 11:14










  • The weird thing is that the solution is basically the same as part of OP's code, just done once more.
    – John Zwinck
    Nov 12 '18 at 11:19










  • @coldspeed Not sure, but from dfs I did what I can, need to check with OP to clear the confusion.
    – Sandeep Kadapa
    Nov 12 '18 at 11:28










  • @JohnZwinck Yes, but that is how OP's output.
    – Sandeep Kadapa
    Nov 12 '18 at 11:30










  • @SandeepKadapa Not sure what the confusion is, they've indicated their expected output at the end of the post.
    – coldspeed
    Nov 12 '18 at 12:04



















1














Simply:



dfs.groupby(['Cat', 'Number']).count()


reproduced the below, which works..



>>> dfs.groupby(['Cat', 'Number']).count()

            Email

Cat Number

ab1 1           2

    2           1

ab2 1           3


OR



>>> dfs.groupby(['Cat', 'Number'])['Email'].count()

Cat  Number

ab1  1         2

     2         1

ab2  1         3

Name: Email, dtype: int64





share|improve this answer























  • This won't produce OP's expected output.
    – coldspeed
    Nov 12 '18 at 11:16










  • @coldspeed, it works correctly , see the POST , OP mentioned dfs=df.groupby(['Email', 'Cat'])['Email'].count().reset_index(name='Number')
    – pygo
    Nov 12 '18 at 11:24










  • Regardless of what they mentioned, it isn't their "expected output".
    – coldspeed
    Nov 12 '18 at 11:53











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














Use groupby+size and reset_index:



df1 = dfs.groupby(['Cat','Number']).size().reset_index(name='Count')


Or:



df1 = dfs.groupby(['Cat','Number'])['Email'].value_counts().reset_index(name='Count')





print(df1)

   Cat  Number  Count

0  ab1       1      2

1  ab1       2      1

2  ab2       1      3





share|improve this answer























  • Isn't your answer incorrect?
    – coldspeed
    Nov 12 '18 at 11:14










  • The weird thing is that the solution is basically the same as part of OP's code, just done once more.
    – John Zwinck
    Nov 12 '18 at 11:19










  • @coldspeed Not sure, but from dfs I did what I can, need to check with OP to clear the confusion.
    – Sandeep Kadapa
    Nov 12 '18 at 11:28










  • @JohnZwinck Yes, but that is how OP's output.
    – Sandeep Kadapa
    Nov 12 '18 at 11:30










  • @SandeepKadapa Not sure what the confusion is, they've indicated their expected output at the end of the post.
    – coldspeed
    Nov 12 '18 at 12:04
















4














Use groupby+size and reset_index:



df1 = dfs.groupby(['Cat','Number']).size().reset_index(name='Count')


Or:



df1 = dfs.groupby(['Cat','Number'])['Email'].value_counts().reset_index(name='Count')





print(df1)

   Cat  Number  Count

0  ab1       1      2

1  ab1       2      1

2  ab2       1      3





share|improve this answer























  • Isn't your answer incorrect?
    – coldspeed
    Nov 12 '18 at 11:14










  • The weird thing is that the solution is basically the same as part of OP's code, just done once more.
    – John Zwinck
    Nov 12 '18 at 11:19










  • @coldspeed Not sure, but from dfs I did what I can, need to check with OP to clear the confusion.
    – Sandeep Kadapa
    Nov 12 '18 at 11:28










  • @JohnZwinck Yes, but that is how OP's output.
    – Sandeep Kadapa
    Nov 12 '18 at 11:30










  • @SandeepKadapa Not sure what the confusion is, they've indicated their expected output at the end of the post.
    – coldspeed
    Nov 12 '18 at 12:04














4












4








4






Use groupby+size and reset_index:



df1 = dfs.groupby(['Cat','Number']).size().reset_index(name='Count')


Or:



df1 = dfs.groupby(['Cat','Number'])['Email'].value_counts().reset_index(name='Count')





print(df1)

   Cat  Number  Count

0  ab1       1      2

1  ab1       2      1

2  ab2       1      3





share|improve this answer














Use groupby+size and reset_index:



df1 = dfs.groupby(['Cat','Number']).size().reset_index(name='Count')


Or:



df1 = dfs.groupby(['Cat','Number'])['Email'].value_counts().reset_index(name='Count')





print(df1)

   Cat  Number  Count

0  ab1       1      2

1  ab1       2      1

2  ab2       1      3






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 12 '18 at 11:16

























answered Nov 12 '18 at 11:12









Sandeep KadapaSandeep Kadapa

6,092429




6,092429












  • Isn't your answer incorrect?
    – coldspeed
    Nov 12 '18 at 11:14










  • The weird thing is that the solution is basically the same as part of OP's code, just done once more.
    – John Zwinck
    Nov 12 '18 at 11:19










  • @coldspeed Not sure, but from dfs I did what I can, need to check with OP to clear the confusion.
    – Sandeep Kadapa
    Nov 12 '18 at 11:28










  • @JohnZwinck Yes, but that is how OP's output.
    – Sandeep Kadapa
    Nov 12 '18 at 11:30










  • @SandeepKadapa Not sure what the confusion is, they've indicated their expected output at the end of the post.
    – coldspeed
    Nov 12 '18 at 12:04


















  • Isn't your answer incorrect?
    – coldspeed
    Nov 12 '18 at 11:14










  • The weird thing is that the solution is basically the same as part of OP's code, just done once more.
    – John Zwinck
    Nov 12 '18 at 11:19










  • @coldspeed Not sure, but from dfs I did what I can, need to check with OP to clear the confusion.
    – Sandeep Kadapa
    Nov 12 '18 at 11:28










  • @JohnZwinck Yes, but that is how OP's output.
    – Sandeep Kadapa
    Nov 12 '18 at 11:30










  • @SandeepKadapa Not sure what the confusion is, they've indicated their expected output at the end of the post.
    – coldspeed
    Nov 12 '18 at 12:04
















Isn't your answer incorrect?
– coldspeed
Nov 12 '18 at 11:14




Isn't your answer incorrect?
– coldspeed
Nov 12 '18 at 11:14












The weird thing is that the solution is basically the same as part of OP's code, just done once more.
– John Zwinck
Nov 12 '18 at 11:19




The weird thing is that the solution is basically the same as part of OP's code, just done once more.
– John Zwinck
Nov 12 '18 at 11:19












@coldspeed Not sure, but from dfs I did what I can, need to check with OP to clear the confusion.
– Sandeep Kadapa
Nov 12 '18 at 11:28




@coldspeed Not sure, but from dfs I did what I can, need to check with OP to clear the confusion.
– Sandeep Kadapa
Nov 12 '18 at 11:28












@JohnZwinck Yes, but that is how OP's output.
– Sandeep Kadapa
Nov 12 '18 at 11:30




@JohnZwinck Yes, but that is how OP's output.
– Sandeep Kadapa
Nov 12 '18 at 11:30












@SandeepKadapa Not sure what the confusion is, they've indicated their expected output at the end of the post.
– coldspeed
Nov 12 '18 at 12:04




@SandeepKadapa Not sure what the confusion is, they've indicated their expected output at the end of the post.
– coldspeed
Nov 12 '18 at 12:04













1














Simply:



dfs.groupby(['Cat', 'Number']).count()


reproduced the below, which works..



>>> dfs.groupby(['Cat', 'Number']).count()

            Email

Cat Number

ab1 1           2

    2           1

ab2 1           3


OR



>>> dfs.groupby(['Cat', 'Number'])['Email'].count()

Cat  Number

ab1  1         2

     2         1

ab2  1         3

Name: Email, dtype: int64





share|improve this answer























  • This won't produce OP's expected output.
    – coldspeed
    Nov 12 '18 at 11:16










  • @coldspeed, it works correctly , see the POST , OP mentioned dfs=df.groupby(['Email', 'Cat'])['Email'].count().reset_index(name='Number')
    – pygo
    Nov 12 '18 at 11:24










  • Regardless of what they mentioned, it isn't their "expected output".
    – coldspeed
    Nov 12 '18 at 11:53
















1














Simply:



dfs.groupby(['Cat', 'Number']).count()


reproduced the below, which works..



>>> dfs.groupby(['Cat', 'Number']).count()

            Email

Cat Number

ab1 1           2

    2           1

ab2 1           3


OR



>>> dfs.groupby(['Cat', 'Number'])['Email'].count()

Cat  Number

ab1  1         2

     2         1

ab2  1         3

Name: Email, dtype: int64





share|improve this answer























  • This won't produce OP's expected output.
    – coldspeed
    Nov 12 '18 at 11:16










  • @coldspeed, it works correctly , see the POST , OP mentioned dfs=df.groupby(['Email', 'Cat'])['Email'].count().reset_index(name='Number')
    – pygo
    Nov 12 '18 at 11:24










  • Regardless of what they mentioned, it isn't their "expected output".
    – coldspeed
    Nov 12 '18 at 11:53














1












1








1






Simply:



dfs.groupby(['Cat', 'Number']).count()


reproduced the below, which works..



>>> dfs.groupby(['Cat', 'Number']).count()

            Email

Cat Number

ab1 1           2

    2           1

ab2 1           3


OR



>>> dfs.groupby(['Cat', 'Number'])['Email'].count()

Cat  Number

ab1  1         2

     2         1

ab2  1         3

Name: Email, dtype: int64





share|improve this answer














Simply:



dfs.groupby(['Cat', 'Number']).count()


reproduced the below, which works..



>>> dfs.groupby(['Cat', 'Number']).count()

            Email

Cat Number

ab1 1           2

    2           1

ab2 1           3


OR



>>> dfs.groupby(['Cat', 'Number'])['Email'].count()

Cat  Number

ab1  1         2

     2         1

ab2  1         3

Name: Email, dtype: int64






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 12 '18 at 11:23

























answered Nov 12 '18 at 11:12









pygopygo

2,4081618




2,4081618












  • This won't produce OP's expected output.
    – coldspeed
    Nov 12 '18 at 11:16










  • @coldspeed, it works correctly , see the POST , OP mentioned dfs=df.groupby(['Email', 'Cat'])['Email'].count().reset_index(name='Number')
    – pygo
    Nov 12 '18 at 11:24










  • Regardless of what they mentioned, it isn't their "expected output".
    – coldspeed
    Nov 12 '18 at 11:53


















  • This won't produce OP's expected output.
    – coldspeed
    Nov 12 '18 at 11:16










  • @coldspeed, it works correctly , see the POST , OP mentioned dfs=df.groupby(['Email', 'Cat'])['Email'].count().reset_index(name='Number')
    – pygo
    Nov 12 '18 at 11:24










  • Regardless of what they mentioned, it isn't their "expected output".
    – coldspeed
    Nov 12 '18 at 11:53
















This won't produce OP's expected output.
– coldspeed
Nov 12 '18 at 11:16




This won't produce OP's expected output.
– coldspeed
Nov 12 '18 at 11:16












@coldspeed, it works correctly , see the POST , OP mentioned dfs=df.groupby(['Email', 'Cat'])['Email'].count().reset_index(name='Number')
– pygo
Nov 12 '18 at 11:24




@coldspeed, it works correctly , see the POST , OP mentioned dfs=df.groupby(['Email', 'Cat'])['Email'].count().reset_index(name='Number')
– pygo
Nov 12 '18 at 11:24












Regardless of what they mentioned, it isn't their "expected output".
– coldspeed
Nov 12 '18 at 11:53




Regardless of what they mentioned, it isn't their "expected output".
– coldspeed
Nov 12 '18 at 11:53


















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Surfing

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Surfing From Wikipedia, the free encyclopedia Jump to navigation Jump to search This article is about stand-up ocean surfing. For other uses, see Surfing (disambiguation). "Surfer" redirects here. For other uses, see Surfer (disambiguation). Surfing A surfer at the Cayucos Pier, Cayucos, California Highest governing body World Surf League (WSL) Characteristics Mixed gender Yes, separate competitions Presence Country or region Worldwide Olympic Will debut in 2020 Surfing on the Gold Coast, Queensland, Australia Surfing is a surface water sport in which the wave rider, referred to as a surfer, rides on the forward or deep face of a moving wave, which usually carries the surfer towards the shore. Waves suitable for surfing are primarily found in the ocean, but can also be found in lakes or rivers in the form of a standing wave or tidal bore. However, surfers can also utilize artifi...
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