Is there a way to use the sort() method or any other method to sort a list by column? Lets say I have the list:
[
[John,2],
[Jim,9],
[Jason,1]
]
You can use list.sort
with its optional key parameter and a lambda expression:
>>> lst = [
... ['John',2],
... ['Jim',9],
... ['Jason',1]
... ]
>>> lst.sort(key=lambda x:x[1])
>>> lst
[['Jason', 1], ['John', 2], ['Jim', 9]]
>>>
This will sort the list in-place.
Note that for large lists, it will be faster to use operator.itemgetter instead of a lambda
:
>>> from operator import itemgetter
>>> lst = [
... ['John',2],
... ['Jim',9],
... ['Jason',1]
... ]
>>> lst.sort(key=itemgetter(1))
>>> lst
[['Jason', 1], ['John', 2], ['Jim', 9]]
>>>
sorted(list, key=lambda x: x[1])
Note: this works on time variable too.
below solution worked for me in case of required number is float. Solution:
table=sorted(table,key=lambda x: float(x[5]))
for row in table[:]:
Ntable.add_row(row)
'
You can use the sorted method with a key.
sorted(a, key=lambda x : x[1])
The optional key
parameter to sort
/sorted
is a function. The function is called for each item and the return values determine the ordering of the sort
>>> lst = [['John', 2], ['Jim', 9], ['Jason', 1]]
>>> def my_key_func(item):
... print("The key for {} is {}".format(item, item[1]))
... return item[1]
...
>>> sorted(lst, key=my_key_func)
The key for ['John', 2] is 2
The key for ['Jim', 9] is 9
The key for ['Jason', 1] is 1
[['Jason', 1], ['John', 2], ['Jim', 9]]
taking the print
out of the function leaves
>>> def my_key_func(item):
... return item[1]
This function is simple enough to write "inline" as a lambda function
>>> sorted(lst, key=lambda item: item[1])
[['Jason', 1], ['John', 2], ['Jim', 9]]
Yes. The sorted
built-in accepts a key
argument:
sorted(li,key=lambda x: x[1])
Out[31]: [['Jason', 1], ['John', 2], ['Jim', 9]]
note that sorted
returns a new list. If you want to sort in-place, use the .sort
method of your list (which also, conveniently, accepts a key
argument).
or alternatively,
from operator import itemgetter
sorted(li,key=itemgetter(1))
Out[33]: [['Jason', 1], ['John', 2], ['Jim', 9]]
Read more on the python wiki.