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I found I had to do a mapping twice to get what I wanted. The first map call 'x' is a A[]. But I need to change each A into a B so I mapped x and returned a new B using the y variable in the second map. Is there a better way? .map(x => x.map(y => <B>{ foo: bar }))

Colin
– Colin

2017-11-07 17:07:13 +00:00
Commented Nov 7, 2017 at 17:07
I think you can return also .map(x => x.map(y => { foo: bar })) as B[] which is maybe more readable

martin
– martin

2017-11-07 21:46:08 +00:00
Commented Nov 7, 2017 at 21:46

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