I have a list like this
data = [('£3.46', 'I001'), ('£10.46', 'I002')]
I want to sort this list with
data.sort()
however the sorted list puts £10.46 before £3.46, I'm assuming this is because the data is a string and £1 comes before £3.
I have tried looking at lambda functions but I can't get my head around it.
One option is to strip "£" and convert the number to float as a sorting key:
data.sort(key=lambda x: float(x[0].lstrip('£')))
print(data)
Output:
[('£3.46', 'I001'), ('£10.46', 'I002')]
You could achieve that in this simple way:
data = [('£10.46', 'I002'), ('£3.46', 'I001')]
def sort_tuple(item):
return float(item[0][1:])
print(sorted(data, key=sort_tuple))
Output:
[('£3.46', 'I001'), ('£10.46', 'I002')]
Process finished with exit code 0
This is a general approach with some limitations, but should work in a wide variety of similar situations.
import re
def float_then_text(values):
'''Convert [str, str, ...] to [(float, str), (float, str), ...]'''
return [(float(re.sub(r'[^\d.]+', '', text)), text) for
for text in values]
data = [('£3.46', 'I001'), ('£10.46', 'I002')]
data.sort(key=float_then_text)
The function float_then_text will convert from ('£3.46', 'I001') to [(3.46, '£3.46'), (1.0, 'I001')] and so on.
Including the text in the keys is optional, but ensures that values with different units get their units sorted in the same order each time as well.
ValueError propagate... where a mix of text/None/float escape - the sort's going to fail anywaydata.sort(key=float_then_text).
sort(key=...) can't sort on a generator, so you have to convert to a sortable quantity like a list first. (Notice that float_then_text is a generator using yield.) Your first point is quite an issue with my answer, and I'm refactoring the answer now to address it)float_then_text a generator... make it a function that takes a single value and returns a single value...