Groovy二维数组 [英] Groovy 2d arrays
问题描述
我有3个数组, l1 [,,,,],l2 [,,,,]和l3 [,,,,]
。每一个都有5个字符。
例如。 l1 [A,B,C,D,E]
2d数组是由这些元素组成的
$ p $ screen = [l1 [],l2 [],l3 []]
所以它看起来是这样的:
screen = [[,,,,],[,,,,],[,,,,]]
我如何迭代这个数组?
我是否调用 screen [5]
?或屏幕[l1 [5]]
?
我可以:
for(i in 1..15){
println screen [i]
}
$ c
$ b $ p请帮忙!!!
如何迭代数组并调用不同的部分,例如每个子数组中的 1st元素
?
谢谢
列表l1 = ['A','B ','C','D','E']
List l2 = ['a','b','c','d','e']
List l3 = [' 1,2,3,4,5]
/ pre>
要迭代所有元素,您可以这样做:
3times {y - >
5.times {x - >
println screen [y] [x]
}
}
或者,您可以使用 mod
和 intdiv
来计算x和y的位置:
15.times {p - >
def x = p%5
def y = p.intdiv(5)
println screen [y] [x]
}
或者,如果您想要每个子列表中的第一个元素,您可以这样做:
println screen * .head()
您需要每个子列表中的第三个元素:
println screen * .getAt(2)
或者,您可以将所有项目放入一维列表中:
def inlineScreen = [* l1,* l2,* l3]
然后通过访问:
15.times {p - >
println inlineScreen [p]
}
或:
def x = 1,y = 1
println inlineScreen [x + y * 5]
或者,如果总是希望将所有元素放在一起(即,总是希望所有数组中的第二个元素放在一起),那么重要的是,那么你可以这样做:
// groupedScreen == [['A','a','1'],[ 'B','b','2'],['C','c','3'],['D','d','4'],['E','e' '5']]
def groupedScreen = [l1,l2,l3] .transpose()
然后,获得第三个元素,你可以这样做:
groupedScreen [2]
I have 3 arrays, l1[,,,,], l2[,,,,] and l3[,,,,]
. Each of which has 5 characters in it.
e.g. l1["A","B","C","D","E"]
The 2d array is made up of these
screen = [l1[],l2[],l3[]]
so it will look as so:
screen = [[,,,,],[,,,,],[,,,,]]
How can I iterate through this array?
do I call screen[5]
? or screen[l1[5]]
?
can I:
for (i in 1..15){
println screen[i]
}
please help!!!
How can I iterate in the array and call different parts, e.g. the 1st element
of each sub array?
Thanks
解决方案 So, given:
List l1 = ['A', 'B', 'C', 'D', 'E']
List l2 = ['a', 'b', 'c', 'd', 'e']
List l3 = ['1', '2', '3', '4', '5']
List screen = [ l1, l2, l3 ]
To iterate all the elements, you could do:
3.times { y ->
5.times { x ->
println screen[ y ][ x ]
}
}
Or, you could use mod
and intdiv
to work out x and y positions:
15.times { p ->
def x = p % 5
def y = p.intdiv( 5 )
println screen[ y ][ x ]
}
Or, if you want the first element from each sub-list, you can do:
println screen*.head()
Or, say you want the 3rd element from each sub-list:
println screen*.getAt( 2 )
Alternatively, you could put all the items into a one-dimensional List:
def inlineScreen = [ *l1, *l2, *l3 ]
Then access then by:
15.times { p ->
println inlineScreen[ p ]
}
or:
def x = 1, y = 1
println inlineScreen[ x + y * 5 ]
Or, if it is important that you always want to get the elements together (ie, you always want the second elements from all the arrays together), then you could do:
// groupedScreen == [['A', 'a', '1'], ['B', 'b', '2'], ['C', 'c', '3'], ['D', 'd', '4'], ['E', 'e', '5']]
def groupedScreen = [ l1, l2, l3 ].transpose()
Then, to get the 3rd elements, you can just do:
groupedScreen[ 2 ]
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