python:我怎么知道我什么时候进入最后一个循环 [英] python: how do i know when i am on the last for cycle

问题描述

for i in range(len(results_histogram)):如果 i!=len(results_histogram)-1:url+=str(results_histogram[i])+','

我的 if 语句正在检查我是否在最后一个循环中，但它不起作用.我做错了什么?

解决方案

为了稍微避免这个问题，你似乎重写了 str.join:

','.join(results_histogram)

如果你得到类似TypeError: sequence item 0: expected string, int found这样的错误，那么你可以用

将中间结果转换成字符串

','.join(map(str, results_histogram))

<小时>

str.join 无疑比在循环中连接多个字符串更有效，因为在 Python 中，字符串是不可变的，所以每次连接都会导致创建一个新字符串，然后必须稍后收集垃圾.

<小时>

特别地，您的示例不起作用"，因为当您只想跳过添加逗号时，您完全跳过了最后一个元素.举个小例子就很清楚了:

<预><代码>>>>x = [1,2,3]>>>对于范围内的 i(len(x)):...如果我 != len(x) - 1:... 打印 str(x[i]) + ',',...1, 2,

所以你可以重写你的例子

for i in range(len(results_histogram)):url += str(results_histogram[i])如果 i!=len(results_histogram)-1:网址 += ','

但你仍然应该坚持使用 str.join.

for i in range(len(results_histogram)):
    if i!=len(results_histogram)-1:
      url+=str(results_histogram[i])+','

my if statement is checking whether i am on the last loop, but it is not working. what am i doing wrong?

解决方案

To avoid the question slightly, you seem to have rewritten str.join:

','.join(results_histogram)

If you get an error like TypeError: sequence item 0: expected string, int found, then you can convert the intermediate results to a string with

','.join(map(str, results_histogram))

---

str.join is undoubtedly more efficient than concatenating multiple strings in a loop, because in Python, strings are immutable, so every concatenation results in the creation of a new string, which then has to be garbage collected later.

---

Specifically, your example is "not working" because you skip the last element entirely, when you only want to skip adding the comma. This is clear and obvious with a small example:

>>> x = [1,2,3]
>>> for i in range(len(x)):
...   if i != len(x) - 1:
...     print str(x[i]) + ',',
... 
1, 2,

So you could rewrite your example as

for i in range(len(results_histogram)):
    url += str(results_histogram[i])
    if i!=len(results_histogram)-1:
      url += ','

But you should still stick with str.join.

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