I have:
d = [1,'q','3', None, 'temp']
I want to replace None value to 'None' or any string
expected effect:
d = [1,'q','3', 'None', 'temp']
a try replace in string and for loop but I get error:
TypeError: expected a character buffer object
7 Answers
Use a simple list comprehension:
['None' if v is None else v for v in d]
Demo:
>>> d = [1,'q','3', None, 'temp']
>>> ['None' if v is None else v for v in d]
[1, 'q', '3', 'None', 'temp']
Note the is None test to match the None singleton.
answered Oct 14, 2013 at 15:36
Martijn Pieters
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3 Comments
joseville
Just a note to reader in case it's not obvious that this creates a new list - it doesn't modify the existing list
d.
alphabetasoup
This can be simplified to
[v or 'None' for v in d] if you know that you don't have empty strings or anything else that evaluates as False (which may be a risky assumption).
Martijn Pieters
@alphabetasoup: exactly, which is why I don't do that in a generic answer on SO.
You can simply use map and convert all items to strings using the str function:
map(str, d)
#['1', 'q', '3', 'None', 'temp']
If you only want to convert the None values, you can do:
[str(di) if di is None else di for di in d]
answered Oct 14, 2013 at 15:44
Saullo G. P. Castro
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Comments
Starting Python 3.6 you can do it in shorter form:
d = [f'{e}' for e in d]
hope this helps to someone since, I was having this issue just a while ago.
2 Comments
uniquegino
what does f'{e}' mean here?
Using a lengthy and inefficient however beginner friendly for loop it would look like:
d = [1,'q','3', None, 'temp']
e = []
for i in d:
if i is None: #if i == None is also valid but slower and not recommended by PEP8
e.append("None")
else:
e.append(i)
d = e
print d
#[1, 'q', '3', 'None', 'temp']
Only for beginners, @Martins answer is more suitable in means of power and efficiency
4 Comments
Martijn Pieters
Don't use
== None; None is a singleton, using is None is faster and more pythonic and PEP-8 compliant.
Steven Rumbalski
Always use
is None if that is what you mean. i can be a type that implements comparison to None as True.
Grijesh Chauhan
@MartijnPieters What is means by
singleton ?Martijn Pieters
An object of which there is only one copy during the lifetime of the program.
None, True, False, Ellipsis and NotImplemented are all singletons part of Python itself.List comprehension is the right way to go, but in case, for reasons best known to you, you would rather replace it in-place rather than creating a new list (arguing the fact that python list is mutable), an alternate approach is as follows
d = [1,'q','3', None, 'temp', None]
try:
while True:
d[d.index(None)] = 'None'
except ValueError:
pass
>>> d
[1, 'q', '3', 'None', 'temp', 'None']
Comments
I did not notice any answer using lambda..
Someone who wants to use lambda....(Look into comments for explanation.)
#INPUT DATA
d =[1,'q','3', None, 'temp']
#LAMBDA TO BE APPLIED
convertItem = lambda i : i or 'None'
#APPLY LAMBDA
res = [convertItem(i) for i in d]
#PRINT OUT TO TEST
print(res)
Comments
My solution was to match for None and replace it.
def fixlist(l: list):
for i, v in enumerate(l):
if v is None:
l[i] = 0
Noneinto a string for whatever you're doing next (that expects a character buffer object), why don't you need to turn the1into a string as well?