I want to sort short list like:
# we can have only 3 types of value: any string numeric value like '555', 'not found' and '' (can have any variation with these options)
row = ['not found', '', '555']
to
# numeric values first, 'not found' less prioritize and '' in the end
['555', 'not found', '']
I trying use
row.sort(key=lambda x: str(x).isnumeric() and not bool(x))
but it's not working
How can I sort it? (numeric values first, 'not found' less prioritize and '' in the end)
def custom_sort(list):
L1 = []
L2 = []
L3 = []
for element in list:
if element.isnumeric():
L1.append(element)
if element == 'Not found':
L2.append(element)
else : L3.append(element)
L1.sort()
L1.append(L2).append(L3)
return L1
L1.append(L2).append(L3). Anyway, I think what you are actually trying to do here is L1.extend(L2); L1.extend(L3).
Edit: Sorting non numeric values per ask
ar = [i for i in row if not i.isnumeric()]
ar.sort(reverse=True)
row = [i for i in row if i.isnumeric()] + ar
'not found' should outrank '' in the sorted list.
This will do the trick too:
row = ['not found', '', 555, 1, '5' , 444]
print(row)
def func(x):
if str(x).isnumeric():
return 1/-int(x) # ordering numerics event if they are strings
elif str(x) == 'not found':
return 2
elif str(x) == '':
return 3
row2 = row.sort(key=func)
print(row)
Results:
['not found', '', 555, 1, '5', 444]
[1, '5', 444, 555, 'not found', '']
This will sort your list and give 'not found' a higher priority than '':
l = [int(a) for a in row if a.isnumeric()] # Or float(a)
l.sort()
row = [str(a) for a in l] +\
['not found'] * row.count('not found') +\
[''] * row.count('')
sortsupports, yet you somehow expectsortto do what you want. You need to define your own key function to produce the priorities as you define them. Alternately, filter the list into the three groupings you specify, sort each grouping, and concatenate the results.