内部列表中的第一项尽可能有效 [英] First items in inner list efficiently as possible

问题描述

我在python A[row,col,value]中有一个协调的存储列表，用于存储非零值.

I have a coordinated storage list in python A[row,col,value] for storing non-zeros values.

如何获取所有行索引的列表?我希望此A[0:][0]可以像print A[0:]一样打印整个列表，但是print A[0:][0]仅打印A[0].

How can I get the list of all the row indexes? I expected this A[0:][0] to work as print A[0:] prints the whole list but print A[0:][0] only prints A[0].

我问的原因是为了有效计算每行 i.e 中的非零值的数量，在range(0,n)上进行迭代，其中n是总行数.比我目前的for i in range(0,n): for j in A: ...方法要便宜得多.

The reason I ask is for efficient calculation of the number of non-zero values in each row i.e iterating over range(0,n) where n is the total number of rows. This should be much cheaper than my current way of for i in range(0,n): for j in A: ....

类似的东西:

c = []
# for the total number of rows
for i in range(0,n):
     # get number of rows with only one entry in coordinate storage list
     if A[0:][0].count(i) == 1: c.append(i)                
return c

结束:

c = []
# for the total number of rows 
for i in range(0,n):
    # get the index and initialize the count to 0 
    c.append([i,0])
    # for every entry in coordinate storage list 
    for j in A:
        # if row index (A[:][0]) is equal to current row i, increment count  
        if j[0] == i:
           c[i][1]+=1
return c

使用Junuxx的答案，此问题和这篇文章我想出了以下(用于返回单身行的数量):对于我当前的问题大小A，它比我最初的尝试要快得多.但是，它仍然随着行和列的数量而增长.我想知道是否有可能不必遍历A而是仅迭代到n吗?

Using Junuxx's answer, this question and this post I came up with the following (for returning the number of singleton rows) which is much faster for my current problems size of A than my original attempt. However it still grows with the number of rows and columns. I wonder if it's possible to not have to iterate over A but just upto n?

# get total list of row indexes from coordinate storage list
row_indexes = [i[0] for i in A]
# create dictionary {index:count}
c = Counter(row_indexes)    
# return only value where count == 1 
return [c[0] for c in c.items() if c[1] == 1]

推荐答案

这应该做到:

c = [x[0] for x in A]

这是一个列表理解，它接受A每个元素的第一个(子)元素.

It's a list comprehension that takes the first (sub-)element of every element of A.

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