Following is my code which is a set of tuples:
以下是我的代码,它是一组元组:
data = {('A',20160129,36.44),('A',20160201,37.37),('A',20160104,41.06)};
print(data);
Output: set([('A', 20160129, 36.44), ('A', 20160104, 41.06), ('A', 20160201, 37.37)])
输出:设置([(‘A’,20160129,36.44),(‘A’,20160104,41.06),(‘A’,20160201,37.37)])
How do I append another tuple ('A', 20160000, 22)
to data
?
如何将另一个元组(‘A’,20160000,22)追加到数据?
Expected output: set([('A', 20160129, 36.44), ('A', 20160104, 41.06), ('A', 20160201, 37.37), ('A', 20160000, 22)])
预期产量:SET([(‘A’,20160129,36.44),(‘A’,20160104,41.06),(‘A’,20160201,37.37),(‘A’,20160000,22)])
Note: I found a lot of resources to append data to a set but unfortunately none of them have the input data in the above format. I tried append
, |
& set
functions as well.
注意:我找到了很多资源来将数据追加到集合中,但不幸的是,它们都没有上述格式的输入数据。我还尝试了APPEND、|和SET函数。
更多回答
data.add(('A', 20160000, 22))?
Data.add((‘A’,20160000,22))?
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the trick is to send it inside brackets so it doesn't get exploded
诀窍是把它放在括号里,这样它就不会爆炸
data.update([('A', 20160000, 22)])
just use data.add
. Here's an example:
只需使用data.add.下面是一个例子:
x = {(1, '1a'), (2, '2a'), (3, '3a')}
x.add((4, '4a'))
print(x)
Output: {(3, '3a'), (1, '1a'), (2, '2a'), (4, '4a')}
输出:{(3,‘3a’),(1,‘1a’),(2,‘2a’),(4,‘4a’)}
s = {('a',1)}
s.add(('b',2))
output: {('a', 1),('b', 2)}
s.update(('c',3))
output: {('a', 1), 3, 'c', ('b', 2)}
There can be loss of generality by using update.
Better to use add.
It is a cleaner way of operating sets.
使用UPDATE可能会失去通用性。最好使用Add。这是一种更清洁的操作设备的方式。
This failure also matching the question title:
此失败还与问题标题匹配:
your_set = set
your_set.add(('A', 20160000, 22))
because it should be:
因为它应该是:
your_set = set()
更多回答
Do you mind explaining why passing in a list of tuple(s) works?
你介意解释一下为什么传递一个元组列表(S)有效吗?
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