Take the following data-frame:
x = np.tile(np.arange(3),3)
y = np.repeat(np.arange(3),3)
df = pd.DataFrame({"x": x, "y": y})
x y
0 0 0
1 1 0
2 2 0
3 0 1
4 1 1
5 2 1
6 0 2
7 1 2
8 2 2
I need to sort it by x first, and only second by y:
df2 = df.sort(["x", "y"]) x y
0 0 0
3 0 1
6 0 2
1 1 0
4 1 1
7 1 2
2 2 0
5 2 1
8 2 2
How can I change the index such that it is ascending again. I.e. how do I get this:
x y
0 0 0
1 0 1
2 0 2
3 1 0
4 1 1
5 1 2
6 2 0
7 2 1
8 2 2
I have tried the following. Unfortunately, it doesn"t change the index at all:
df2.reindex(np.arange(len(df2.index)))
Update index after sorting data-frame repeat: Questions
Create list of single item repeated N times
5 answers
I want to create a series of lists, all of varying lengths. Each list will contain the same element e, repeated n times (where n = length of the list).
How do I create the lists, without using a list comprehension [e for number in xrange(n)] for each list?
Answer #1
You can also write:
[e] * n
You should note that if e is for example an empty list you get a list with n references to the same list, not n independent empty lists.
Performance testing
At first glance it seems that repeat is the fastest way to create a list with n identical elements:
>>> timeit.timeit("itertools.repeat(0, 10)", "import itertools", number = 1000000)
0.37095273281943264
>>> timeit.timeit("[0] * 10", "import itertools", number = 1000000)
0.5577236771712819
But wait - it"s not a fair test...
>>> itertools.repeat(0, 10)
repeat(0, 10) # Not a list!!!
The function itertools.repeat doesn"t actually create the list, it just creates an object that can be used to create a list if you wish! Let"s try that again, but converting to a list:
>>> timeit.timeit("list(itertools.repeat(0, 10))", "import itertools", number = 1000000)
1.7508119747063233
So if you want a list, use [e] * n. If you want to generate the elements lazily, use repeat.
What is the best way to repeatedly execute a function every x seconds?
5 answers
I want to repeatedly execute a function in Python every 60 seconds forever (just like an NSTimer in Objective C). This code will run as a daemon and is effectively like calling the python script every minute using a cron, but without requiring that to be set up by the user.
In this question about a cron implemented in Python, the solution appears to effectively just sleep() for x seconds. I don"t need such advanced functionality so perhaps something like this would work
while True:
# Code executed here
time.sleep(60)
Are there any foreseeable problems with this code?
Answer #1
If your program doesn"t have a event loop already, use the sched module, which implements a general purpose event scheduler.
import sched, time
s = sched.scheduler(time.time, time.sleep)
def do_something(sc):
print("Doing stuff...")
# do your stuff
s.enter(60, 1, do_something, (sc,))
s.enter(60, 1, do_something, (s,))
s.run()
If you"re already using an event loop library like asyncio, trio, tkinter, PyQt5, gobject, kivy, and many others - just schedule the task using your existing event loop library"s methods, instead.
Answer #2
Lock your time loop to the system clock like this:
import time
starttime = time.time()
while True:
print "tick"
time.sleep(60.0 - ((time.time() - starttime) % 60.0))
