Python running threads sequentially

Question

I'm experimenting with running code parallel in Python.

My use case is as follows: I have a process that I need to run thousands of times. But due to it's dependency on a download I need to be careful how many I run at the same time (I have a limited number of requests each minute). I therefore would like to always run the function 10 times simultaneously, each time one is finished I would like to start up a new process.

After doing some quick reading the threading module seems to be what I'm looking for. However, some small experiments seem to show that Python does not run the threads in paralell.

When I run the following code

import threading
import time

wait = 0.4

def f(x):
 for j in range(1,100):
    time.sleep(wait)
    print(x)


threading.Thread(target = f(1)).start()
threading.Thread(target = f(2)).start()

I get in the commandline

Not a singe 2 shows up untill all 100 ones are printed out.

How does this come? In theory f(1) and f(2) should be running simultaneously right?

you're calling the functions when creating the threads. Classic dupe — Jean-François Fabre, Sep 24 '18 at 12:56
Also note: `range(1, 100)` will print 99 times, not 100; the second number is the limit at which it stops before. — Ctrl S, Sep 24 '18 at 13:00

score 9 · Accepted Answer · answered Sep 24 '18 at 12:56

9

This line

threading.Thread(target = f(1)).start()

basically says "start a new thread and execute the result of f(1) in it".

What you want is "start a new thread and execute f with argument 1 in it", which translates to

threading.Thread(target = f, args=(1,)).start()

answered Sep 24 '18 at 12:56

MSpiller

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Note also that `f(1)` returns `None`, which happens to be a valid `target` ("`target` is the callable object to be invoked by the `run()` method. **Defaults to `None`, meaning nothing is called.**" - emphasis mine) – Amadan Sep 24 '18 at 12:59

Python running threads sequentially

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