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sql select group by a having count(1) > 1 equivalent in python pandas?


sql select group by a having count(1) > 1 equivalent in python pandas?

By : user2956733
Date : November 22 2020, 03:03 PM
this will help Instead of writing email_cnt[email_cnt.size > 1], just write email_cnt[email_cnt > 1] (there's no need to call.size again). This uses the Boolean series email_cnt > 1 to return only the relevant values of email_cnt.
For example:
code :
>>> customers = pd.DataFrame({'Email':['foo','bar','foo','foo','baz','bar'],
                              'CustomerID':[1,2,1,2,1,1]})
>>> email_cnt = customers.groupby('Email')['CustomerID'].size()
>>> email_cnt[email_cnt > 1]
Email
bar      2
foo      3
dtype: int64


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Python - Pandas - Plotting Count of Column by Group - Graphing Each Group Over Time

Python - Pandas - Plotting Count of Column by Group - Graphing Each Group Over Time


By : user2215976
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further Sorry if the title is horribly vague, its hard to express the issue in a few words.
code :
d = {'level' : ['ERROR', 'ERROR', 'ERROR', 'ERROR', 'ERROR', 'ERROR', 'ERROR', 'ERROR', 'ERROR', 'ERROR'], 
 'DATE' : ['2014-07-29 12:35:55.916', '2014-07-29 12:35:55.916', '2014-07-29 12:35:55.916', '2014-07-29 12:35:55.874', '2014-07-29 12:35:55.908', '2014-07-29 12:35:55.908', '2014-07-29 12:35:55.908', '2014-07-29 12:35:55.908', '2014-07-29 12:35:55.908', '2014-07-29 12:35:55.975'],
 'APP' : ['app1', 'app1', 'app1', 'app2', 'app3', 'app3', 'app3', 'app3', 'app3', 'app4']}
df = pd.DataFrame(d)
df.pivot_table(values='level', index='DATE', columns='APP', aggfunc=len).plot(kind='bar')
R equivalent of SQL SELECT COUNT(*) ... GROUP BY

R equivalent of SQL SELECT COUNT(*) ... GROUP BY


By : user3422764
Date : March 29 2020, 07:55 AM
this will help I'm trying to find how to count the number of integers of each type in a vector. Eg, how many 1, 2, and 3 are there (without hard-coding == 1,2,3): , aggregate is very handy in this situation
code :
> aggregate(data.frame(count = test_vec), list(value = test_vec), length)

  value count
1     1    10
2     2     7
3     3     4
With Pandas in Python, select only the rows where group by group count is 1

With Pandas in Python, select only the rows where group by group count is 1


By : Pramod Mali
Date : March 29 2020, 07:55 AM
Hope that helps I've filtered my data as suggested here: With Pandas in Python, select the highest value row for each group , Easier
code :
df.groupby('author').filter(lambda x: len(x)==1)


     author        cat  val
id                         
0   author1  category2   15
1   author2  category4    9
Is there a tidyverse equivalent to SELECT...COUNT(*)...GROUP BY...?

Is there a tidyverse equivalent to SELECT...COUNT(*)...GROUP BY...?


By : Yash MATHUR
Date : March 29 2020, 07:55 AM
it helps some times For an introduction to these basic operations in the tidyverse, I'd suggest reading Wickham and Grolemund's excellent R for Data Science in the first instance: http://r4ds.had.co.nz/
You can use dplyr and magrittr packages to do the following in an easy to follow way:
code :
# Install the tidyverse
library(tidyverse)

# Create data
place = rep(c('AL','AK','AZ','AR','CA','CO','CT','DE','FL','GA','HI'), times=4)
measure = rep(c('meas1','meas2','meas3','meas4'), each=11)
set.seed(200)
rating = sample(c('good','bad'), size = 44, prob=c(2,1), replace=T)
df = data.frame(place, measure, rating)

# Do some analysis
df %>% 
  group_by(place) %>% 
  summarise(mean_score = mean(rating == "good"), n = n()) %>% 
  arrange(desc(mean_score))
pandas equivalent select count(distinct col1, col2) group by col3

pandas equivalent select count(distinct col1, col2) group by col3


By : Arsen
Date : March 29 2020, 07:55 AM
will help you Make DataFrame: , drop_duplicates with groupby + count
code :
(df.drop_duplicates()
   .groupby('Site_Where_Served')
   .Site_Where_Served.count()
   .reset_index(name='Site_Visit_Count')
)

  Site_Where_Served  Site_Visit_Count
0          hospital                 3
1         inpatient                 1
(df[['Person', 'Service_Date']]
   .apply(tuple, 1)
   .groupby(df.Site_Where_Served)
   .nunique()
   .reset_index(name='Site_Visit_Count')
)

  Site_Where_Served  Site_Visit_Count
0          hospital                 3
1         inpatient                 1
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