如何将元组转换为不带逗号和括号的值字符串 [英] How to transform a tuple to a string of values without comma and parentheses

问题描述

我使用 sql 查询检索数据

bounds = cursor.fetchone()

然后我得到一个元组:

(34.2424, -64.2344, 76.3534, 45.2344)

我想要一个像 34.2424 -64.2344 76.3534 45.2344

这样的字符串

是否存在可以做到这一点的函数?

解决方案

使用 str.join():

<预><代码>>>>mystring = ' '.join(map(str, (34.2424, -64.2344, 76.3534, 45.2344)))>>>打印我的字符串34.2424 -64.2344 76.3534 45.2344

您必须在此处使用 map(它将元组中的所有项目转换为字符串)，否则您将收到 TypeError.

<小时>

对 map() 的一些说明 功能:

map(str, (34.2424, -64.2344, 76.3534, 45.2344) 等价于 [str(i) for i in (34.2424, -64.2344, 76.3534, 45)]/代码>.

它比使用列表推导式要快一点:

$ python -m timeit "map(str, (34.2424, -64.2344, 76.3534, 45.2344))"1000000 个循环，最好的 3 个:每个循环 1.93 微秒$ python -m timeit "[str(i) for i in (34.2424, -64.2344, 76.3534, 45.2344)]"100000 个循环，最好的 3 个:每个循环 2.02 微秒

如对此答案的评论所示，str.join() 可以使用生成器而不是列表.通常，这会更快，但在这种情况下，它慢.

如果我要这样做:

' '.join(itertools.imap(str, (34.2424, -64.2344, 76.3534, 45.2344)))

它会比使用 map() 慢.区别在于 imap() 返回一个生成器，而 map() 返回一个列表(在 python 3 中它返回一个生成器)

如果我要这样做:

''.join(str(i) for i in (34.2424, -64.2344, 76.3534, 45.2344))

由于在here列表理解中加括号要慢>.

<小时>

在您(OP)的情况下，任何一个选项都无关紧要，因为性能在这里似乎没什么大不了的.但是，如果您曾经处理过大的浮点数/整数元组，那么现在您知道如何使用以实现最高效率:)

I retrieved data from a sql query by using

bounds = cursor.fetchone()

And I get a tuple like:

(34.2424, -64.2344, 76.3534, 45.2344)

And I would like to have a string like 34.2424 -64.2344 76.3534 45.2344

Does a function exist that can do that?

解决方案

Use str.join():

>>> mystring = ' '.join(map(str, (34.2424, -64.2344, 76.3534, 45.2344)))
>>> print mystring
34.2424 -64.2344 76.3534 45.2344

You'll have to use map here (which converts all the items in the tuple to strings) because otherwise you will get a TypeError.

---

A bit of clarification on the map() function:

map(str, (34.2424, -64.2344, 76.3534, 45.2344) is equivalent to [str(i) for i in (34.2424, -64.2344, 76.3534, 45.2344)].

It's a tiny bit faster than using a list comprehension:

$ python -m timeit "map(str, (34.2424, -64.2344, 76.3534, 45.2344))"
1000000 loops, best of 3: 1.93 usec per loop
$ python -m timeit "[str(i) for i in (34.2424, -64.2344, 76.3534, 45.2344)]"
100000 loops, best of 3: 2.02 usec per loop

As shown in the comments to this answer, str.join() can take a generator instead of a list. Normally, this would be faster, but in this case, it is slower.

If I were to do:

' '.join(itertools.imap(str, (34.2424, -64.2344, 76.3534, 45.2344)))

It would be slower than using map(). The difference is that imap() returns a generator, while map() returns a list (in python 3 it returns a generator)

If I were to do:

''.join(str(i) for i in (34.2424, -64.2344, 76.3534, 45.2344))

It would be slower than putting brackets around the list comprehension, because of reasons explained here.

---

In your (OP's) case, either option does not really matter, as performance doesn't seem like a huge deal here. But if you are ever dealing with large tuples of floats/integers, then now you know what to use for maximum efficiency :).

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