I have a list/array like so:
{{{1}, {2}, {3}, {4}, {5}, {6}}, {{7}, {8}, {9}, {10}, ... }}
I want to remove the curly brackets around each element (so I can use ListDensityPlot) so that I have:
{{1, 2, 3, 4, 5, 6}, {7, 8, 9, ...}}
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3 Answers
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list = {{{1}, {2}, {3}, {4}, {5}, {6}}, {{7}, {8}, {9}, {10}}};
Flatten /@ list
{{1, 2, 3, 4, 5, 6}, {7, 8, 9, 10}}
Or use the operator form
Map[Flatten] @ list
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Just to put my comment on record, I propose
{{{1}, {2}, {3}, {4}, {5}, {6}}, {{7}, {8}, {9}, {10}}}[[All, All, 1]]
which gives
{{1, 2, 3, 4, 5, 6}, {7, 8, 9, 10}}
answered Jul 31, 2017 at 17:46
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$\begingroup$ I see that you have adopted my solution since this was last asked. ;^) $\endgroup$ Commented Jul 31, 2017 at 22:19
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1$\begingroup$ @Mr.Wizard. Why not? Your methods are often the best, and I can still learn even at my advanced age. $\endgroup$ Commented Jul 31, 2017 at 23:59
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$\begingroup$ I am sincerely flattered. Also I strongly suspect your memory works better than mine, age notwithstanding. $\endgroup$ Commented Aug 1, 2017 at 9:46
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Given your list as:
list = {{{1}, {2}, {3}, {4}, {5}, {6}}, {{7}, {8}, {9}, {10}}};
Use Flatten to remove the inner brackets.
Flatten /@ list
Which gives:
{{1, 2, 3, 4, 5, 6}, {7, 8, 9, 10}}
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1$\begingroup$ Isn't this the same answer as eldo's above? $\endgroup$– MarcoBCommented Jul 31, 2017 at 19:27
Flatten[{{{1, 2}, {3, 4}, {5, 6}}, {{7, 8}, {9, 10}}}, {{1}, {2, 3}}]. (Pretty sure this is a dupe...) $\endgroup$Flatten /@ {{{1, 2}, {3, 4}, {5, 6}}, {{7, 8}, {9, 10}}}$\endgroup${{{1}, {2}, {3}, {4}, {5}, {6}}, {{7}, {8}, {9}, {10}}}[[All, All, 1]]$\endgroup$