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How can i match fields with wildcards using jq?


By : user2175908
Date : October 14 2020, 02:24 PM
wish help you to fix your issue Filter keys that start with Test and get only the attribute of your choice using the select() expression
code :
jq 'to_entries[] | select(.key|startswith("Task")).value.attributes.processid' json


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How to use prefix wildcards like '*abc' with match-against


By : user3673310
Date : March 29 2020, 07:55 AM
it fixes the issue Match doesn't work with starting wildcards, so matching with *abc* won't work. You will have to use LIKE to achieve this:
code :
SELECT * FROM user WHERE user_login LIKE '%abc';
user_login user_login_rev
xyzabc     cbazyx

Using wildcards in Sql to match patterns


By : user3103082
Date : March 29 2020, 07:55 AM
Does that help im trying to match patterns in a sql using LIKE. , You can make use of regexp_like
code :
WITH DATA AS(
    SELECT 'XKA' str FROM dual UNION ALL
    SELECT 'XKB' FROM dual UNION ALL
    SELECT 'XSA' FROM dual UNION ALL
    SELECT 'XSB' FROM dual UNION ALL
    SELECT 'XAA' FROM dual
    )
    SELECT str
    FROM DATA
    WHERE regexp_like (str, 'X[^KS]');



X -> first character
^[KS] -> second character not in K or S

Pandas count number of rows that match on certain fields but don't match on other fields


By : Mark71
Date : March 29 2020, 07:55 AM
Does that help You could count the number of rows with unique BRTH_DT and ADDRESS using
code :
df.drop_duplicates().groupby(['FRST_NM','LAST_NM'])['ADDRESS'].count()
import numpy as np
import pandas as pd
np.random.seed(2016)
N = 50
df = pd.DataFrame({'FRST_NM':np.random.choice(['Charles','Edgar Allan','Agatha'], N),
                   'LAST_NM':np.random.choice(['Dickens', 'Poe', 'Christie'], N),
                   'BRTH_DT':np.random.choice([1812, 1809, 1890], N),
                   'ADDRESS':np.random.choice(['Landport','Boston','Torquay'], N)})

result = pd.DataFrame(
    {'num_unique' : (df.drop_duplicates()
                       .groupby(['FRST_NM','LAST_NM'])['ADDRESS']
                       .count()), 
     'count' : df.groupby(['FRST_NM', 'LAST_NM'])['ADDRESS'].count()})
result['percent'] = result['num_unique'] / result['count']
print(result)
                      count  num_unique   percent
FRST_NM     LAST_NM                              
Agatha      Christie      4           4  1.000000
            Dickens       8           4  0.500000
            Poe           7           5  0.714286
Charles     Christie      7           6  0.857143
            Dickens       4           4  1.000000
            Poe           9           6  0.666667
Edgar Allan Christie      4           3  0.750000
            Dickens       4           3  0.750000
            Poe           3           3  1.000000

sed - replace match containing two wildcards


By : Ilya Lee
Date : March 29 2020, 07:55 AM
it helps some times All proposed answers work to a degree, but both have problems (a trailing " is wrongly dropped).
code :
OPTIONS="-g"
OPTIONS="-u ntp:ntp -p /var/run/ntpd.pid -g"
OPTIONS="-u ntp:ntp -p /var/run/ntpd.pid -g"
OPTIONS="-u ntp:ntp
OPTIONS="-u"
OPTIONS="-g"
OPTIONS="-u ntp:ntp -p /var/run/ntpd.pid -g"
OPTIONS="-u ntp:ntp -p /var/run/ntpd.pid -g"
OPTIONS="-u ntp:ntp"
OPTIONS="-u"

Can sscanf be used to match wildcards?


By : partik
Date : March 29 2020, 07:55 AM
With these it helps The format "%1d%1d33%1d" should be correct, assuming your inputs are all numbers. But you haven't told us what specific inputs it's failing on. You should consider that the strings "1 2334" and " 1\n\n233 \t 4" would actually match because %d will eat whitespace until it finds an integer.
Beware that if you were to use "%2d33%1d" this would be even worse, because a 2-character integer can be a single digit with a negative.
code :
int matches( const char * s )
{
    return s
        && isdigit(s[0])
        && isdigit(s[1])
        && '3' ==  s[2]
        && '3' ==  s[3]
        && isdigit(s[4]);
}
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