Question

string1 = "The wind, "
string2 = "which had hitherto carried us along with amazing rapidity, "
string3 = "sank at sunset to a light breeze; "
string4 = "the soft air just ruffled the water and "
string5 = "caused a pleasant motion among the trees as we approached the shore, "
string6 = "from which it wafted the most delightful scent of flowers and hay."

我尝试过：

for i in range(6): 
     message +=string(i)

但是它不起作用并显示错误：字符串未定义

我想直接操作var，我知道将它们放在列表中要容易得多，但是想像一下，如果您有1000个字符串，则很难在列表中编写每个字符串。

Answer 1

使用join()：

cList = [string1, string2, string3, string4, string5, string6]

print("".join(cList))

我建议的变量应该是列表中的变量，而不是n个变量：

x = ["The wind, ", "which had hitherto carried us along with amazing rapidity, ", "sank at sunset to a light breeze; ", "the soft air just ruffled the water and ", "caused a pleasant motion among the trees as we approached the shore", "from which it wafted the most delightful scent of flowers and hay."]    

print("".join(x))

单线：

print("".join([string1, string2, string3, string4, string5, string6]))

Answer 2

如果首先可以将字符串放在列表中，那就更好了。否则：

for s in [string1, string2, string3, string4, string5, string6]:
    message += s

Answer 3

您正在使用不同的变量。您将必须分别调用每个变量才能将它们连接起来，因为您正试图像在列表中那样调用它们。尝试将它们添加到数组或列表中。

Answer 4

也许您在寻找eval吗？

 message = ''.join([eval('string'+str(i)) for i in range(1,7)])

Answer 5

只需一口气写下您的故事（请注意字符串之间没有逗号）：

message = ("The wind, "
           "which had hitherto carried us along with amazing rapidity, "
           "sank at sunset to a light breeze; "
           "the soft air just ruffled the water and "
           "caused a pleasant motion among the trees as we approached the shore, "
           "from which it wafted the most delightful scent of flowers and hay.")

或者如果您想输入较少的引号：

import inspect

message = """
          The wind,
          which had hitherto carried us along with amazing rapidity,
          sank at sunset to a light breeze;
          the soft air just ruffled the water and
          caused a pleasant motion among the trees as we approached the shore,
          from which it wafted the most delightful scent of flowers and hay.
          """

message = inspect.cleandoc(message)  # remove unwanted indentation
message = message.replace('\n', ' ')  # remove the newlines

Answer 6

对此问题有多种解决方案，评论中给出了一种。您收到该错误的原因是因为字符串不存在。您正在致电string(i)，您究竟希望这样做吗？

此外，您正在执行的循环逻辑不正确。当进行编码但未获得预期结果时，第一道防线是调试。在这种情况下，请了解您要遍历的内容，本质上是数字。继续并打印该 i 变量，以便您了解发生了什么。您完全可以访问 stringX 变量。它们必须包含在可迭代中，以便您对其进行迭代。更不用说 for循环迭代是错误的，因为 range（x）提供了从 0 到 x-1 ，在您的情况下为 0 1 2 3 4 5 。您将知道，如果您已调试。我必须说，编码的一部分是调试。习惯去做是一件好事。

这里是documentation on Python about strings.

string.join（words [，sep]）将单词的列表或元组与中间出现的sep连接起来。 sep的默认值是单个空格字符。始终是string.join（string.split（s，sep），sep）等于s。

这意味着您可以使用字符串的方法join来连接字符串。该方法要求您向其传递字符串的列表。您的代码如下所示：

string1 = "The wind, " string2 = "which had hitherto carried us along with amazing rapidity, " string3 = "sank at sunset to a light breeze; " string4 = "the soft air just ruffled the water and " string5 = "caused a pleasant motion among the trees as we approached the shore, " string6 = "from which it wafted the most delightful scent of flowers and hay." message = "".join([string1, string2, string3, string4, string5, string6]) print(message)

输出：

The wind, which had hitherto carried us along with amazing rapidity, sank at sunset to a light breeze; the soft air just ruffled the water and caused a pleasant motion among the trees as we approached the shore, from which it wafted the most delightful scent of flowers and hay.

由于我不确定您的目标是什么，因此，为了有所不同，我将假设您有任意数量的字符串变量传递给您。这甚至更容易处理，因为如果定义一个名为 join_strings 的方法，则传递的变量的值已经是一个列表。整洁吧？因此，您的代码将类似于：

def join_strings(*strings): return "".join(strings)

非常矮小又甜蜜，不是吗？然后，您将像这样调用该方法：

join_strings(string1, string2, string3, string4, string5, string6)

很棒的事情是明天您可能只有3个字符串或8个字符串，并且仍然可以使用。当然，知道为什么还要保存这样的字符串会更有用，因为我确信您可以根据需要使用更合适的数据结构（例如使用列表开头）。

下次您发布到StackOverflow时，最好尝试展示一些尝试做的事情来理解您的问题，而不仅仅是粘贴问题。

如何在for循环中连接这些字符串变量？

6 个答案: