如何比较列表中连续整数的最后几位数？

Question

在我创建的程序中，我需要将文件中的整数添加到列表中，然后确定每个整数的最后一位，并将其与下一个整数的最后一位进行比较，然后继续此循环，直到清单已与下列清单进行比较并存储结果。我可以将文件中的整数添加到列表中，并确定每个整数的最后一位，但我无法比较最后的数字。我一直在使用的代码，

with open('test.txt') as f: 
    my_list = [] 
    for line in f: 
      my_list.extend(int(i) for i in line.split()) 

for elem in my_list: 
    nextelem = my_list[my_list.index(elem)-len(my_list)+1] 

one_followed_by_1 = 0 
one_followed_by_2 = 0 
one_followed_by_3 = 0 
one_followed_by_4 = 0 

for elem in my_list: 
    if elem > 9: 
     last_digit = elem % 10 
     last_digit_next = nextelem % 10 
     if last_digit == 1 and last_digit_next == 1: 
      one_followed_by_1 += 1 
     elif last_digit == 1 and last_digit_next == 2: 
      one_followed_by_2 += 1 
     elif last_digit == 1 and last_digit_next == 3: 
      one_followed_by_3 += 1 
     elif last_digit == 1 and last_digit_next == 4: 
      one_followed_by_4 += 1 

print one_followed_by_1 
print one_followed_by_2 
print one_followed_by_3 
print one_followed_by_4 

但是，这不适合我。任何帮助将不胜感激。

Answer 1

你让事情太复杂了。首先，我们可以简单地写这样的解析器：

with open('test.txt') as f: 
    my_list = [int(i) for line in f for i in line.split()] 

下一页，而不是构建nextelem是复杂的方式，我们可以使用zip(my_list,my_list[1:])，遍历当前和下一个项目同时：

for n0,n1 in zip(my_list,my_list[1:]): 
    pass 

当然现在我们仍然需要处理计数。但是，我们可以使用collections库的Counter来做到这一点。像：

from collections import Counter 

ctr = Counter((n0%10,n1%10) for n0,n1 in zip(my_list,my_list[1:])) 

所以我们甚至不需要for循环。现在Counter是一本字典。它映射元组(i,j)上以i结尾的数字数cij，后面跟着一个以j结尾的数字。

例如，打印数字，如：

print ctr[(1,1)] # 1 followed by 1 
print ctr[(1,2)] # 1 followed by 2 
print ctr[(1,3)] # 1 followed by 3 
print ctr[(1,4)] # 1 followed by 4 

或程序在全：

from collections import Counter 

with open('test.txt') as f: 
    my_list = [int(i) for line in f for i in line.split()] 

ctr = Counter((n0%10,n1%10) for n0,n1 in zip(my_list,my_list[1:])) 

print ctr[(1,1)] # 1 followed by 1 
print ctr[(1,2)] # 1 followed by 2 
print ctr[(1,3)] # 1 followed by 3 
print ctr[(1,4)] # 1 followed by 4