增加python

Question

以下程序给出错误IndexError：list index超出范围

newlist=[2,3,1,5,6,123,436,124,223.......,213,213,213,56,2387567,3241,2136]   
# total 5600 values seprated by commas in the above list

emptylist = []
for values in newlist:
          convstr = str(values)
          convstr = convstr.split(",")
          emptylist.extend(convstr)

k=0
for i in range(5600):
    for j in range(0,4):
        print(i,j,emptylist[k])
        k=k+1

但是当我使用相同的程序时，newlist包含1000个值，它可以正常工作

newlist=[2,3,1,5,6,123,436,124,223.......,213]  
 # total 1000 values seprated by commas in the above list

emptylist = []
for values in newlist:
          convstr = str(values)
          convstr = convstr.split(",")
          emptylist.extend(convstr)

k=0
for i in range(1000):
    for j in range(0,4):
        print(i,j,emptylist[k])
        k=k+1

那么为什么它没有使用5600值显示索引超出范围但是它使用了1000个值？

TRIED使用列表中的Len也不起作用

Answer 1

如果您有5600个元素的列表。您将它们转换为不增加其数量的字符串。迭代5600次，在每次迭代中，您将k增加4倍，然后使用k索引到列表中 - 结果：索引错误

newlist=[2,3,1,5,6,123,436,124,223.......,213,213,213,56,2387567,3241,2136]   
# total 5600 values seprated by commas in the above list

emptylist = []
for values in newlist:
    convstr = str(values)  # values is ONE number, its string is also one number
    convstr = convstr.split(",")   # there are no , in numbers but you get a [number]
    emptylist.extend(convstr)      # this adds the string of a int to the list

k=0  # you index by k
for i in range(5600):  # you do this 5600  times
    for j in range(0,4):  # your print AND INCREASE k 4 times
        print(i,j,emptylist[k])
        k=k+1             # after about 5600 / 4 iterations of the loop your k is 
                          # larger then the amount in your list