HelloWorld - 1 year ago 71
Python Question

# How can I append lists with items from another list and leave these empty in a simulation?

Here is my code:

``````import random
listx = []
listy = []
listz = []

#function

def function(r,s,q):
listy=[]
if len(listx)==4 or len(listx)==8:
listy.append((listx, t))
print "y", listy

# here it starts:

for t in range(1,25):

randomnumber = random.uniform(0.0, 1.0)

if randomnumber <= 0.5:
listx.append((t))
print "x", listx

function(5,6,8) #this function generates listy from input of listx

if listy != 0: #if listy is not empy anymore, fill listz with items of listy
listz.append(listy)
del listy
print "z", listz
``````

I have 3 lists;
`listx`
,
`listy`
and
`listz`
. I generate random numbers to
`listx`
. If
`ready==0`
(always) I call the function
`(function(r,s,q))`
. If a condition
`(len==4 or 8)`
is met, the items in the list are appended to listy.

At this point, I would like to add these numbers from
`listy`
to
`listz`
and leave
`listy`
empty again. The items that I transported from
`listy`
to
`listz`
should be removed from
`listx`
.

Does anyone know how to solve this?

I think this may do you want. Changes to your code are indicated with `# UPPERCASE COMMENTS`.

``````import random
listx = []
listy = []
listz = []

#function

def function(r,s,q):

listy=[]
if len(listx) in (4, 8):  # STREAMLINED
listy.append((listx, t))
print "y", listy

# here it starts:

for t in range(1,25):

randomnumber = random.uniform(0.0, 1.0)

if randomnumber <= 0.5:
listx.append((t))
print "x", listx

if ready == 0: # condition, lets say: ready is always 0
function(5,6,8) # this function generates listy from input of listx

#### REWRITTEN
if listy: # not empty?
listz += listy[0][0]  # copy numbers from listy to listz
del listx[:len(listy[0][0])]  # remove numbers transported from listx
del listy  # empty listy
print "z", listz
``````
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