在 Python 中的单独一行上打印变量的每个项目
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Printing each item of a variable on a separate line in Python
提问by MSK
I am trying to print a list in Python that contains digits and when it prints the items in the list all print on the same line.
我正在尝试用 Python 打印一个包含数字的列表，当它打印列表中的所有项目时，它们都打印在同一行上。
print ("{} ".format(ports))
here is my output
这是我的输出
[60, 89, 200]
how can I see the result in this form:
我怎样才能看到这种形式的结果：
60
89
200
I have tried print ("\n".join(ports))but that does not work.
我试过了，print ("\n".join(ports))但这不起作用。
回答by jramirez
ports = [60, 89, 200]
for p in ports:
print (p)
回答by Martijn Pieters
Loop over the list and print each item on a new line:
循环遍历列表并在新行上打印每个项目：
for port in ports:
print(port)
or convert your integers to strings before joining:
或在加入之前将整数转换为字符串：
print('\n'.join(map(str, ports)))
or tell print()to use newlines as separators and pass in the list as separate arguments with the *splat syntax:
或者告诉print()使用换行符作为分隔符，并使用*splat 语法将列表作为单独的参数传递：
print(*ports, sep='\n')
Demo:
演示：
>>> ports = [60, 89, 200]
>>> for port in ports:
... print(port)
...
60
89
200
>>> print('\n'.join(map(str, ports)))
60
89
200
>>> print(*ports, sep='\n')
60
89
200
回答by abarnert
i have tried print ("\n".join(ports)) but does not work.
我试过打印 ("\n".join(ports)) 但不起作用。
You're very close. The only problem is that, unlike print, joindoesn't automatically convert things to strings; you have to do that yourself.
你很亲近。唯一的问题是，与 不同print，join它不会自动将事物转换为字符串；你必须自己做。
For example:
例如：
print("\n".join(map(str, ports)))
… or …
… 或者 …
print("\n".join(str(port) for port in ports))
If you don't understand either comprehensions or map, both are equivalent* to this:
如果您不理解推导式或map，则两者都等同于*：
ports_strs = []
for port in ports:
ports_strs.append(str(port))
print("\n".join(ports_strs))
del ports_strs
In other words, map(str, ports)will give you the list ['60', '89', '200'].
换句话说，map(str, ports)会给你列表['60', '89', '200']。
Of course it would be silly to write that out the long way; if you're going to use an explicit forstatement, you might as well just print(port)directly each time through the loop, as in jramirez's answer.
当然，把它写得很长是愚蠢的。如果您要使用显式for语句，则最好print(port)每次都直接通过循环，如 jramirez 的回答。
* I'm actually cheating a bit here; both give you an iteratorover a sort of "lazy list" that doesn't actually exist with those values. But for now, we can just pretend it's a list. (And in Python 2.x, it was.)
* 我实际上在这里作弊了；两者都为您提供了一个迭代器，用于一种实际上并不存在于这些值的“懒惰列表”。但是现在，我们可以假装它是一个列表。（在 Python 2.x 中，确实如此。）
回答by abarnert
If you are on Python 3.x:
如果您使用的是 Python 3.x：
>>> ports = [60, 89, 200]
>>> print(*ports, sep="\n")
60
89
200
>>>
Otherwise, this will work:
否则，这将起作用：
>>> ports = [60, 89, 200]
>>> for p in ports:
... print p
...
60
89
200
>>>
