How to get id of those persons who have matching month and year only in dob column

How to get id of those persons who have matching month and year only in dob column

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How to get id of those persons who have matching month and year only in dob column
Tag : mysql , By : RinKaMan
Date : November 28 2020, 11:01 PM

To fix the issue you can do In my table, I have some records and they have some duplicate date of birth values, so I want the id of those people who have matching year and month only .
 FROM table group by date having count(*) >= 2 
select YEAR(date) AS 'year', MONTH(date) AS 'month' COUNT(*) c FROM table GROUP BY column HAVING c > 1

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GET Persons Birthday, Days Left Until Bday, and Persons Age this year in MS SQL 2012

Tag : sql , By : noboruwatanabe
Date : March 29 2020, 07:55 AM
With these it helps The script I have below works great for displaying birthdays but it will not display the birthday on the birthday date and also how do I get the day's left for that person's birthday ? and finally how old will the person be on this birthday. , Try this for age:

Index and matching column under year and month

Tag : excel , By : Justin Bowers
Date : March 29 2020, 07:55 AM
wish of those help As per my comment above you can have a unique key to use in the offset formula so that the correct numbers populate
You could create month-year unique key to be used as the lookup array inside the match function

Change date format of pandas column (month-day-year to day-month-year)

Tag : python , By : Sid
Date : March 29 2020, 07:55 AM
I hope this helps . One can initialize the data for the days using strings, then convert the strings to datetimes. A print can then deliver the objects in the needed format.
I will use an other format (with dots as separators), so that the conversion is clear between the steps.
import pandas as pd
data = {'day': ['3-20-2019', None, '2-25-2019'] }
df = pd.DataFrame( data )

df['day'] = pd.to_datetime(df['day'])
df['day'] = df['day'].dt.strftime('%d.%m.%Y')
df[ df == 'NaT' ] = '' 
In [56]: df['day']                                                  
0    3-20-2019
1         None
2    2-25-2019
Name: day, dtype: object
In [58]: df['day']                                               
0   2019-03-20
1          NaT
2   2019-02-25
Name: day, dtype: datetime64[ns]
In [59]: df['day'].dt.strftime('%d.%m.%Y')
0    20.03.2019
1           NaT
2    25.02.2019
Name: day, dtype: object
In [73]: df[ df=='NaT' ] = ''

In [74]: df
0  20.03.2019
2  25.02.2019

How can I get the oldest entry (month-year) using month and year table column (int) in mysql?

Tag : mysql , By : Sebastián Ucedo
Date : March 29 2020, 07:55 AM
this will help I have a table with a column that keeps the year and anoher column that keep the month. How can I get the oldest date (minimun td_month-td_year) grouping by son_id? Considerations: table_id could be not in order , You need to get the min year first:
SELECT y.son_id,y.min_year,min(t.mt_month) AS min_month
FROM mytable t
  select son_id,  min(mt_year) AS min_year from mytable group by son_id) y
ON t.son_id=y.son_id
AND t.mt_year=y.min_year
GROUP BY y.son_id,y.min_year;

Python - Extract year and month from a single column of different year and month arrangements

Tag : python , By : anov
Date : September 26 2020, 01:00 AM
I wish did fix the issue. You can make use of the extract dataframe string method to split the date strings up. Since the year can precede or follow the month, we can get a bit creative and have a Year1 column and Year2 columns for either position. Then use np.where to create a single Year column pulls from each of these other year columns.
For example:
import numpy as np

split_dates = df["Date"].str.extract(r"(?P<Year1>\d+)?-?(?P<Month>\w+)-?(?P<Year2>\d+)?")

split_dates["Year"] = np.where(

split_dates = split_dates[["Year", "Month"]]
  Year Month
0   18   Jan
1   18   Jan
2   18   Feb
3   18   Feb
4   17   Oct
5   17   Oct
pd.merge(df, split_dates, how="inner", left_index=True, right_index=True)
     Date  Quantity Year Month
0  18-Jan      3476   18   Jan
1  18-Jan        20   18   Jan
2  18-Feb       789   18   Feb
3  18-Feb       409   18   Feb
4  Oct-17        81   17   Oct
5  Oct-17       640   17   Oct
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