使用for循环和条件语句搜索python列表

Question

a = [1,1,1,4,4,4,5] 
b = [20150602, 20150603, 20150604, 20150605, 20150606, 20150607,20150608] 
c = zip(a,b) 

output = [] 
for i in range(0, len(c)-1): 
    if c[i][0] == c[i+1][0] and c[i+1][1] - c[i][1] <= 3: 
     output.append(c[i]) 
print output 

此代码使用条件搜索2个列表。如果列表b中的列表a AND（元素n + 1） - （元素n）< = 3中的元素n ==元素n + 1，则它打印结果。我遇到的问题是它切断了一个应该打印的元素。 （即此代码将输出

[(1, 20150602), (1, 20150603), (4, 20150605), (4, 20150606)] 

而应输出

[(1, 20150602), (1, 20150603), (1,20150604), (4, 20150605), (4, 20150606), (4, 20150607)]) 

Answer 1

见第二次追加到输出列表:)

In [14]: 

a = [1,1,1,4,4,4,5] 
b = [20150602, 20150603, 20150604, 20150605, 20150606, 20150607,20150608] 
c = zip(a,b) 

output = [] 
for i in range(0, len(c)-1): 
    if c[i][0] == c[i+1][0] and c[i+1][1] - c[i][1] <= 3: 
     output.append(c[i]) 
     output.append(c[i+1]) 
    output = sorted(list(set(output))) 
print output 



[(1, 20150602), (1, 20150603), (1, 20150604), (4, 20150605), (4, 20150606), (4, 20150607)] 

Answer 2

a = [1,1,1,4,4,4,5] 
b = [20150602, 20150603, 20150604, 20150605, 20150606, 20150607,20150608] 
c = zip(a,b) 

output = [] 
for i in range(0, len(c)-1): 
    if c[i][0] == c[i+1][0] and c[i+1][1] - c[i][1] <= 3: 
     output.append(c[i]) 
    elif c[i][0] != c[i+1][0]: 
     print 'equality failed for numbers: %d and %d' % (c[i][0], c[i+1][0]) 
    elif c[i+1][1] - c[i][1] > 3: 
     print 'Subtraction is greater than three for: %d and %d' % (c[i][1], c[i+1][1]) 

print output 

输出：

equality failed for numbers: 1 and 4 
equality failed for numbers: 4 and 5 
[(1, 20150602), (1, 20150603), (4, 20150605), (4, 20150606)] 

你缺少两个元素的原因是因为当i在for循环2 c[i][0]将返回1.现在，当执行if语句的这一部分：c[i][0] == c[i+1][0]返回false因为c[i][0]返回1和c[i+1][0]回报4，他们不相等。当i等于5时会发生同样的情况。