CODEWITHSUNDEEP
pythonlistfor-loop
Chucks
Chucks

Reputation: 899

python list from for loop into one list

Python list in for loop, how do I have all in one list

import os
dir="/tmp/logvol/logs/fil/sap/archive/"
n=os.listdir(dir)
#print n

for x in n:
 #print x
 l = [];
 l.append(os.path.join(dir,x))
 print l

output:

['/tmp/logvol/logs/fil/sap/archive/m.log']
['/tmp/logvol/logs/fil/sap/archive/k.log']
['/tmp/logvol/logs/fil/sap/archive/j.log']
['/tmp/logvol/logs/fil/sap/archive/p.log']

Instead of multiple lists, I need all in one list.

['/tmp/logvol/logs/fil/sap/archive/m.log', '/tmp/logvol/logs/fil/sap/archive/k.log', '/tmp/logvol/logs/fil/sap/archive/j.log', '/tmp/logvol/logs/fil/sap/archive/p.log']

Upvotes: 2

Views: 167

Answers (3)

oleg
oleg

Reputation: 4182

You recreate l list every iteration of for loop

for x in n:
    #print x
    l = []; # here You recreate list every iteration
    l.append(os.path.join(dir,x))
    print l

to avoid this instantiate list before for loop

l = []; # here You recreate list every iteration
for x in n:
    l.append(os.path.join(dir,x))
print l

and pick other name for variable, not l because http://www.python.org/dev/peps/pep-0008/#names-to-avoid

Upvotes: 0

Jeff
Jeff

Reputation: 7210

Another way of doing it is with list comprehension.

l = [os.path.join(dir,x) for x in n]

Upvotes: 9

mdml
mdml

Reputation: 22912

You need to move your declaration of the list l outside the loop:

l = []
for x in n:
 #print x
 l.append(os.path.join(dir,x))

Also, you don't need semicolons in Python.

Upvotes: 5

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