Question

在这段代码中，我想迭代列表中的每个项目，从索引[4]开始，在索引[3]处完成。有没有办法改进我编写代码来实现这一目标的繁琐方式？我试了几天[4：3]，但这只迭代索引4到6而且从不0-3;

mar = int(input())
days = ["0","1","2","3","4","5","6"]
count = 0
x = "0"
while count < mar:
  for a in days[4],[5],[6],[0],[1],[2],[3]:
    if count < mar:
      count += 1
      x = a   
print(x)

Answer 1

itertools islice + cycle

您可以使用itertools创建迭代器。这是一个懒惰的解决方案，需要您在指定范围内调用迭代器上的next。

from itertools import islice, cycle

c = islice(cycle(days), 5, None)

for _ in range(len(days)):
    print(next(c))

5
6
0
1
2
3
4

仅推荐for small iterables：

请注意，此工具包的成员可能需要大量辅助存储（取决于可迭代的长度）。

Answer 2

试试这个：

square[square.columns == col.name]

编辑：既然您现在告诉我们您的目的，我认为您所需要的只是this。

Answer 3

非常感谢您的评论和建议，它完美无缺。我正在考虑类似于你的建议，即创建两个列表，所以我很高兴看到作为python编程的新手我正在思考正确的方法。请在下面将您的建议集成到我的原始代码中，以实现我的目标;

 #This programme is intended to find the number of day of week for N-th day (mar) of a year provided that in this year January 1 is based on user entry (start_day). Days of week are numbered as: 0 — Sunday, 1 — Monday, 2 — Tuesday, ..., 6 — Saturday. An integer K in the range 1 to 365 is given.                        
start_day = int (input("Enter day number to start:- ")) 
mar = int(input("Enter nth day of the year to find:-"))    
count = 0      
days = ["0","1","2","3","4","5","6"]
days_before = []
while count < mar:
# Loop                
  for day in days: 

    if int(day) > start_day: 
      if count < mar:
        count += 1     # Either print
        print (day)    
    else:
        days_before.append(day)   # Or dump to second list
  else:                             # When finished
    for day in days_before: 
      if count < mar:
        count += 1       # Print second list
        print (day)

Answer 4

start = 4
for i in range(start, start+min(mar, len(days))):
    print(days[i % len(days)])

# mar = 4 --> 4, 5, 6, 0
# mar = 10 --> 4, 5, 6, 0, 1, 2, 3

改进代码，迭代python中列表中的每个项目，从0

4 个答案:

itertools islice + cycle