Reputation: 1
My code is assigning values to names out of a list. I need to know how to retrieve or set a default value if there is not an entry in my list to assign to one of the names. My code so far:
array =[[a],[b],[c],[d]]
no1 = array[0]
no2 =array[1]
no3 =array[2]
no4 =array[3]
# if array[4] exists:
no5 = array[4]
else
no5 = 0
I tried-
try:
array[4]
except ValueError:
no5 = 0
but it came up as array[4] out of range.
Just to clarify, as my code writing isn't spectacular, basically I am being given three different inputs for a program which should read these inputs and then write the outputs to a file. The problem is that all of the inputs have varying array lengths. So what I'm trying to do is somehow get the program to check if there is an entry for (say) array[4] and if that entry doesn't exist, create entry array[4] and make it equal to zero.
Upvotes: 0
Views: 969
Reputation: 87376
You are trying to catch the wrong exception type
try:
no5 = array[4]
except IndexError:
no5 = 0
A nifty one-line solution for this is to use itertools
import itertools
(no1,no2,no3,no4,no5),_ = zip(*itertools.izip_longest(array[:5],range(5),fillvalue=0))
Upvotes: 0
Reputation: 18570
Why not something like the following?
if len(array) > 4:
no5 = array[4]
else:
no5 = 0
Upvotes: 0
Reputation: 19766
instead of try
you can also check the length of the array
def: elem_exists(arr, idx):
return 0 <= ldx < len(arr)
Upvotes: 0
Reputation: 184141
You should be catching IndexError
(the actual error being raised by array[4]
) rather than ValueError
.
Upvotes: 1
Reputation: 47978
That you went to try
and / except
is good. If you got an IndexError
then you're really close, you just caught the wrong type of error in your except
. This would've worked:
try:
no5 = array[4]
except IndexError:
no5 = 0
Upvotes: 1