Iterating through rows in Pandas

Question

I am having an issue applying some maths to my dataframe

Current df:

name	lastConnected	check
test1	1647609274746	Connection
test2	1647609274785	Connection
test3	1647000000000	Connection
test4	1647609274756	Connection

Desired df: i now want to create a new column and check if server is still online

name	lastConnected	check
test1	1647609274746	Connection
test2	1647609274785	Connection
test3	1647000000000	No connection
test4	1647609274756	Connection

current code:

def checkServer():
        for index, row in df.iterrows():
                timeNow = int((time.time_ns) // 1000000)
                lastSeenTime = row['lastConnected']
                timeDifference = currentTime - lastSeenTime
                if timeDifference > 5000:
                        df['check'] = "No connection"
                else:
                        df['check'] = "Connection"
        return df

My issue:

As you can see in my current dataframe it gives Connection to them all even though test3 should have No connection. From my troubleshooting, i printed the timeDifference into each row and i got the same time difference even though its all got different times. As a result I think my for loop might be the issue.

Use this site to get current time in miliseconds: currentmillis.com

Where am I going wrong?

So i want to get the current timestamp in miliseconds and compare to the lastConnected time. If over 5000 then say No connection else connected. E.g currentmillis.com Current timestamp: 1647612255595 If 1647612255595 (current timestamp)- 1647609274746(timestamp for test1) is greater than 500 then check = No connection — Ali987
– Ali987, Commented Mar 18, 2022 at 14:03

Corralien · Accepted Answer · 2022-03-18 14:20:09Z

2

Maybe you expect:

df['check'] = np.where(pd.Timestamp.today().timestamp() * 1000 - df['lastConnected'] > 5000,
                       'No connection', 'Connection')
print(df)

# Output
    name  lastConnected          check
0  test1  1647609274746     Connection
1  test2  1647609274785     Connection
2  test3  1647000000000  No connection
3  test4  1647609274756     Connection

Old answer

Use np.where:

df['check'] = np.where(df['lastConnected'].diff().abs().gt(5000),
                       'No connection', 'Connection')
print(df)

# Output
    name  lastConnected          check
0  test1        1647609     Connection
1  test2        1647579     Connection
2  test3        1640009  No connection

edited Mar 18, 2022 at 14:20

answered Mar 18, 2022 at 13:29

Corralien

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1 Comment

Corralien Over a year ago

I updated my answer. Can you check it please?

mozway · Accepted Answer · 2022-03-18 13:29:08Z

2

IIUC, don't use iterrows but a vectorial function:

N = 5000
df.loc[df['lastConnected'].diff().lt(-N), 'check'] = 'No connection'

or to create the column from sratch:

N = 5000
df['check'] = np.where(df['lastConnected'].diff().lt(-N),
                       'No connection', 'connection')

output:

    name  lastConnected          check
0  test1        1647609     Connection
1  test2        1647579     Connection
2  test3        1640009  No connection

answered Mar 18, 2022 at 13:29

mozway

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