为什么内置连接对我的代码没有影响?

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我有一个错误，我减少到这个:

a = ['a','b','c']打印(之前"，一个)" ".join(a)打印(之后"，一个)

输出这个:

runfile('C:/program.py', wdir=r'C:/')在 ['a', 'b', 'c'] 之前['a', 'b', 'c'] 之后

这是怎么回事?

解决方案

str.join 不能就地操作，因为字符串对象在 Python 中是不可变的.相反，它返回一个全新的字符串对象.

如果你想让 a 引用这个新对象，你需要显式地重新赋值:

a = " ".join(a)

演示:

>>>a = ['a','b','c']>>>打印之前"，一个在 ['a', 'b', 'c'] 之前>>>a = " ".join(a)>>>打印之后"，一个在 a b c 之后>>>

I had a bug that I reduced down to this:

a = ['a','b','c'] print( "Before", a ) " ".join(a) print( "After", a )

Which outputs this:

runfile('C:/program.py', wdir=r'C:/') Before ['a', 'b', 'c'] After ['a', 'b', 'c']

What's going on here?

解决方案

str.join does not operate in-place because string objects are immutable in Python. Instead, it returns an entirely new string object.

If you want a to reference this new object, you need to explicitly reassign it:

a = " ".join(a)

Demo:

>>> a = ['a','b','c'] >>> print "Before", a Before ['a', 'b', 'c'] >>> a = " ".join(a) >>> print "After", a After a b c >>>