I have a list of variable length and am trying to find a way to test if the list item currently being evaluated is the longest string contained in the list. And I am using Python 2.6.1
For example:
mylist = ['abc','abcdef','abcd']
for each in mylist:
if condition1:
do_something()
elif ___________________: #else if each is the longest string contained in mylist:
do_something_else()
Surely there’s a simple list comprehension that’s short and elegant that I’m overlooking?
Answers:
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Method 1
From the Python documentation itself, you can use max:
>>> mylist = ['123','123456','1234'] >>> print max(mylist, key=len) 123456
Method 2
def longestWord(some_list):
count = 0 #You set the count to 0
for i in some_list: # Go through the whole list
if len(i) > count: #Checking for the longest word(string)
count = len(i)
word = i
return ("the longest string is " + word)
or much easier:
max(some_list , key = len)
Method 3
What should happen if there are more than 1 longest string (think ’12’, and ’01’)?
Try that to get the longest element
max_length,longest_element = max([(len(x),x) for x in ('a','b','aa')])
And then regular foreach
for st in mylist:
if len(st)==max_length:...
Method 4
len(each) == max(len(x) for x in myList) or just each == max(myList, key=len)
Method 5
To get the smallest or largest item in a list, use the built-in min and max functions:
lo = min(L) hi = max(L)
As with sort, you can pass in a “key” argument that is used to map the list items before they are compared:
lo = min(L, key=int) hi = max(L, key=int)
http://effbot.org/zone/python-list.htm
Looks like you could use the max function if you map it correctly for strings and use that as the comparison. I would recommend just finding the max once though of course, not for each element in the list.
Method 6
def LongestEntry(lstName):
totalEntries = len(lstName)
currentEntry = 0
longestLength = 0
while currentEntry < totalEntries:
thisEntry = len(str(lstName[currentEntry]))
if int(thisEntry) > int(longestLength):
longestLength = thisEntry
longestEntry = currentEntry
currentEntry += 1
return longestLength
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0