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Setting flag column depending on whether column contains a given string


Setting flag column depending on whether column contains a given string

By : user2185063
Date : November 20 2020, 04:01 AM
seems to work fine Can anyone see why this isn't working? , You have to remove list, need only string:
code :
df.loc[df['Name'].str.contains('Andy'),'Andy'] = 1
df.loc[df['Name'].str.contains('Andy|George'),'Andy'] = 1


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Setting colours to FlexiGrid rows depending on column value or attribute href column value

Setting colours to FlexiGrid rows depending on column value or attribute href column value


By : user2984671
Date : March 29 2020, 07:55 AM
it helps some times If you really dont want " " (A space) in div within td, Just try this code.
code :
$('tr:has(td > div:empty)').css("color","red");
How do you create a flag that takes the value of column value depending whether the values make up the majority of the c

How do you create a flag that takes the value of column value depending whether the values make up the majority of the c


By : user3635366
Date : March 29 2020, 07:55 AM
wish of those help I have a table that looks like this...
code :
WITH
  summary As
(
  SELECT
    your_table.*,
    COUNT(*) OVER (PARTITION BY id)  AS id_count,
    COUNT(*) OVER (PARTITION BY id, city)  AS id_city_count
  FROM
    your_table
)
SELECT
  summary.*,
  MAX(
    CASE WHEN id_city_count * 2 > id_count THEN city ELSE NULL END
  ) 
  OVER (PARTITION BY id)
FROM
  summary
Setting column equal to value depending on another column pandas

Setting column equal to value depending on another column pandas


By : Nicholas Vanderweit
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I am stuck on how to set the value of solvent column in each row to a number in the num column in the data frame. eg I need num to be equal to 9 when solvent is Nonane and num equal to 8 when solvent is Octane etc. Any help would be great. , use .loc with a boolean mask
code :
df.loc[df['solvent'] == 'NONANE', 'num'] = 9
df.loc[df['solvent'] == 'OCTANE', 'num'] = 8
d = {'NONANE':9, 'OCTANE':8, 'HEPTANE':7, 'HEXANE':6}
df['num'] = df['solvent'].map(d)
Get count of rows and a flag depending on some column values

Get count of rows and a flag depending on some column values


By : Ronin
Date : March 29 2020, 07:55 AM
may help you . You need to use aggregate function COUNT() to count all rows based from familyID and MAX() with CASE statement to check if membertype column has main:
code :
SELECT  familyID,
        COUNT(*) AS "Count",
        MAX(CASE WHEN memberType = 'main' THEN 1 ELSE 0 END) AS hasMain
FROM    TableName
GROUP   BY familyID
Add new column in pandas dataframe using empty string or the value from column A depending on the value on column B

Add new column in pandas dataframe using empty string or the value from column A depending on the value on column B


By : Alize
Date : March 29 2020, 07:55 AM
Hope that helps just use .loc with pandas conditions to assign just the rows you need:
code :
df.loc[df['price_if_0005'] == 0, 'label'] = df['price']
import pandas as pd
from io import StringIO

s = """
         price |   tpo_count | tpo             |   price_if_0005 
   0 |  1.4334 |           1 | n               |          0.0004 
   1 |  1.4335 |           1 | n               |          0      
   2 |  1.4336 |           1 | n               |          0.0001 
   3 |  1.4337 |           1 | n               |          0.0002 
   4 |  1.4338 |           1 | n               |          0.0003 
   5 |  1.4339 |           1 | n               |          0.0004 
   6 |  1.434  |           1 | n               |          0      
   7 |  1.4341 |           1 | n               |          0.0001 
   8 |  1.4342 |           3 | noq             |          0.0002 
   9 |  1.4343 |           3 | noq             |          0.0003 
  10 |  1.4344 |           3 | noq             |          0.0004 """

df = pd.read_csv(StringIO(s), sep="\s+\|\s+")
df.loc[df['price_if_0005'] == 0, 'label'] = df['price']
df['label'].fillna('',inplace=True)
print(df)
     price  tpo_count  tpo  price_if_0005   label
0   1.4334          1    n         0.0004        
1   1.4335          1    n         0.0000  1.4335
2   1.4336          1    n         0.0001        
3   1.4337          1    n         0.0002        
4   1.4338          1    n         0.0003        
5   1.4339          1    n         0.0004        
6   1.4340          1    n         0.0000   1.434
7   1.4341          1    n         0.0001        
8   1.4342          3  noq         0.0002        
9   1.4343          3  noq         0.0003        
10  1.4344          3  noq         0.0004        
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