在一分钟内运行 python function 但精确到新分钟的 00 秒

Question

I want to run the code exact at 1 min timeframe but only when second will be zero.

import yfinance as yf
company_name = '^NSEI'

# need to loop this code
df = yf.Ticker(company_name).history(period='1d', interval='1m')
print(df.tail(1))

If I run it multiple times, I am getting output as,

2022-03-04 09:49:00+05:30  16283.849609  16283.849609  16269.049805  16279.150391       0          0             0
2022-03-04 09:50:00+05:30  16277.700195  16278.650391  16268.150391  16268.150391       0          0             0
2022-03-04 09:51:00+05:30  16269.200195  16274.799805  16262.950195  16274.549805       0          0             0
2022-03-04 09:52:00+05:30  16274.299805  16274.299805  16263.400391  16263.799805       0          0             0
2022-03-04 09:53:23+05:30  16263.250000  16263.250000  16263.250000  16263.250000       0          0             0

You can see last row, and its printed at "09:53:23+05:30" How can I run this code after exact 1 min timeframe? If I use time.sleep(60) then it may sleep for 60 seconds but I will not get output exact at "00" seconds.

I want to loop this so that it print rows perfectly after 1 min and thus rows should be printed as 09:52:00+05:30, 09:53:00+05:30, 09:54:00+05:30, etc

Answer 1

If you want the code to wait for an exact minute and run the code when the second becomes 0, you can increase the precision for which the code is sleeping (for eg wait 100th of a second)

import datetime as dt
import time

while True:
    while dt.datetime.now().second:
        time.sleep(0.01)
    print(dt.datetime.now()) # Your yfinance code here

But if you just want the output to show 00 as seconds, then you can round it down as pointed out by @k33da_the_bug in the comments.