I have list:
test_list = ['one','two',None]
Is there any simple way to replace None with 'None' ,without using index, because index for None maybe different each time.
I tried :
conv = lambda i : i or 'None'
res = [conv(i) for i in test_list]
It works ,is there another way to do so ?
In this way all the data types would be converted to string
test_list = ['one','two',None]
res = [str(i) for i in test_list]
In this the data type will also be preserved
res = ['None' if i is None else i for i in test_list]
Another way:
test_list = ['one', 'two', None]
res = [i or 'None' for i in test_list]
Even better if you're working with numbers, since int(None) gives an error:
test_list = [1, 2, None]
res = [i or 0 for i in test_list]
['one', ''] to ['one', 'None'], or to ['one', 0], respectively, which is probably not desired.
['one', 0] can't happen, since I said you'll use [i or 0 for i in test_list] only if you will use actual numbers, not strings
None with 'None', which is the question. So this answer is just wrong.
conv = lambda i : i or 'None'and use that expression in the list comprehension?i or 'None'can be too broad. It will turn more than justNoneinto a string.''and0would also be converted to string'None'.