如何在python中的列表列表中比较项目

Question

I am a newbie to python and just learning things as I do my project and here I have a list of lists which I need to compare between the second and last column and get the output for the one which has the largest distance. Moreover, I am replicating a list. If someone could help me to do it within a single list of lists in ag it could be really helpful. Thanks in advance

if this is an INPUT then the output should be,

ag = [['chr12','XX',1,5,4],
     ['chr12','XX',2,5,3],
     ['chr13','ZZ',6,10,4],
     ['chr13','ZZ',8,9,1],
     ['ch14','YY',12,15,3],['ch14','YY',12,15,3]]

EXPECTED OUTPUT:

     ['chr12','XX',1,5,4]
     ['chr13','ZZ',6,10,4]
     ['ch14','YY',12,15,3]

#However I tried of replicating the list like
#INPUT
ag = 
 [['chr12','XX',1,5,4],
 ['chr12','XX',2,5,3],
 ['chr13','ZZ',6,10,4],
 ['chr13','ZZ',8,9,1],
 ['ch14','YY',12,15,3],
 ['ch14','YY',12,15,3]]
 bg = 
 [['chr12','XX',1,5,4],
 ['chr12','XX',2,5,3],
 ['chr13','ZZ',6,10,4],
 ['chr13','ZZ',8,9,1],
 ['ch14','YY',12,15,3],
 ['ch14','YY',12,15,3]]


#The code which I tried was

c= []
for i in ag:
 for j in bg:
    if i[0]==j[0] and i[1]==j[1] and i[4]>j[4]:
        c.append(i)


the output which i get is
[['chr12', 'XX', 1, 5, 4], ['chr13', 'ZZ', 6, 10, 4]]

Answer 1

In short : To compare items in an iterable (eg list) of lists, use the keyword agument key of the max/min function. It takes a function or lambda expression and compares the given values by the result of the key function when given each value.

---

Assuming that what you really want is to reduce a list of lists so that the entries' second elements are unique and that you want the last elements to determin which entry to keep in case of redundant 2nd values:

If there is any problem regarding iteration, itertools has the answer. In this case, we just need the groupby method and standard Python's max method with the keyword argument key .

from itertools import groupby

def filter(matrix):
    filtered = [] # create a result list to hold the rows (lists) we want
    for key, rows in groupby(matrix, lambda row: row[1]): # get the rows grouped by their 2nd element and iterate over that
        filtered.append(max(rows, key=lambda row: row[-1])) # add the line that among its group has the largest last value to our result
    return filtered # return the result

We could squeeze this into a single generator expression or list comprehension but for a beginner, the above code should be complex enought.

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Answer 2

Your list ag and bg are completely duplicating, so I give this example for your issue. Hope it help.

>>>ag = [['chr12','XX',1,5,4],['chr12','XX',2,5,3],['chr13','ZZ',6,10,4],['chr13','ZZ',8,9,1],['ch14','YY',12,15,3]]
>>>bg = [['chr12','XX',1,5,4],['chr12','XX',2,5,3],['chr13','ZZ',6,10,4],['chr13','ZZ',8,9,1]]
>>>[i for i in ag + bg if i not in ag or i not in bg]
[['ch14', 'YY', 12, 15, 3]]