I want to extract a specific element from the list and add it
5
$begingroup$
{p[0, 8] -> 0.538464, p[1, 7] -> 0.107693, p[1, 8] -> 0.161539,
p[2, 6] -> 0.0323078, p[2, 7] -> 0.0484617, p[2, 8] -> 0.0538464,
p[3, 5] -> 0.00969234, p[3, 6] -> 0.0145385, p[3, 7] -> 0.0161539,
p[4, 4] -> 0.0029077, p[4, 5] -> 0.00436155, p[4, 6] -> 0.00484617,
p[5, 3] -> 0.000872311, p[5, 4] -> 0.00130847, p[5, 5] -> 0.00145385,
p[6, 2] -> 0.000261693, p[6, 3] -> 0.00039254, p[6, 4] -> 0.000436155,
p[7, 1] -> 0.000078508, p[7, 2] -> 0.000117762, p[7, 3] -> 0.000130847,
p[8, 0] -> 0.0000235524, p[8, 1] -> 0.0000353286, p[8, 2] -> 0.000039254,
p[9, 0] -> 0.0000176643, p[9, 1] -> 0.0000117762, p[10, 0] -> 8.83215*10^-6}
I want to extract a specific element from the list and add it.
For example, I want to extract the element which becomes 8 by the element of p [i, j] and add it.
p[0, 8] + p[1, 7] + p[2, 6] + ... = answer
Please advise an efficient way, thank you.
list-manipulation
edited Jan 8 at 6:21
Αλέξανδρος Ζεγγ
4,46011029
asked Jan 8 at 5:37
Kenta KawaiKenta Kawai
385
$endgroup$
add a comment |
5
$begingroup$
{p[0, 8] -> 0.538464, p[1, 7] -> 0.107693, p[1, 8] -> 0.161539,
p[2, 6] -> 0.0323078, p[2, 7] -> 0.0484617, p[2, 8] -> 0.0538464,
p[3, 5] -> 0.00969234, p[3, 6] -> 0.0145385, p[3, 7] -> 0.0161539,
p[4, 4] -> 0.0029077, p[4, 5] -> 0.00436155, p[4, 6] -> 0.00484617,
p[5, 3] -> 0.000872311, p[5, 4] -> 0.00130847, p[5, 5] -> 0.00145385,
p[6, 2] -> 0.000261693, p[6, 3] -> 0.00039254, p[6, 4] -> 0.000436155,
p[7, 1] -> 0.000078508, p[7, 2] -> 0.000117762, p[7, 3] -> 0.000130847,
p[8, 0] -> 0.0000235524, p[8, 1] -> 0.0000353286, p[8, 2] -> 0.000039254,
p[9, 0] -> 0.0000176643, p[9, 1] -> 0.0000117762, p[10, 0] -> 8.83215*10^-6}
I want to extract a specific element from the list and add it.
For example, I want to extract the element which becomes 8 by the element of p [i, j] and add it.
p[0, 8] + p[1, 7] + p[2, 6] + ... = answer
Please advise an efficient way, thank you.
list-manipulation
edited Jan 8 at 6:21
Αλέξανδρος Ζεγγ
4,46011029
asked Jan 8 at 5:37
Kenta KawaiKenta Kawai
385
$endgroup$
add a comment |
5
5
5
$begingroup$
{p[0, 8] -> 0.538464, p[1, 7] -> 0.107693, p[1, 8] -> 0.161539,
p[2, 6] -> 0.0323078, p[2, 7] -> 0.0484617, p[2, 8] -> 0.0538464,
p[3, 5] -> 0.00969234, p[3, 6] -> 0.0145385, p[3, 7] -> 0.0161539,
p[4, 4] -> 0.0029077, p[4, 5] -> 0.00436155, p[4, 6] -> 0.00484617,
p[5, 3] -> 0.000872311, p[5, 4] -> 0.00130847, p[5, 5] -> 0.00145385,
p[6, 2] -> 0.000261693, p[6, 3] -> 0.00039254, p[6, 4] -> 0.000436155,
p[7, 1] -> 0.000078508, p[7, 2] -> 0.000117762, p[7, 3] -> 0.000130847,
p[8, 0] -> 0.0000235524, p[8, 1] -> 0.0000353286, p[8, 2] -> 0.000039254,
p[9, 0] -> 0.0000176643, p[9, 1] -> 0.0000117762, p[10, 0] -> 8.83215*10^-6}
I want to extract a specific element from the list and add it.
For example, I want to extract the element which becomes 8 by the element of p [i, j] and add it.
p[0, 8] + p[1, 7] + p[2, 6] + ... = answer
Please advise an efficient way, thank you.
list-manipulation
edited Jan 8 at 6:21
Αλέξανδρος Ζεγγ
4,46011029
asked Jan 8 at 5:37
Kenta KawaiKenta Kawai
385
$endgroup$
{p[0, 8] -> 0.538464, p[1, 7] -> 0.107693, p[1, 8] -> 0.161539,
p[2, 6] -> 0.0323078, p[2, 7] -> 0.0484617, p[2, 8] -> 0.0538464,
p[3, 5] -> 0.00969234, p[3, 6] -> 0.0145385, p[3, 7] -> 0.0161539,
p[4, 4] -> 0.0029077, p[4, 5] -> 0.00436155, p[4, 6] -> 0.00484617,
p[5, 3] -> 0.000872311, p[5, 4] -> 0.00130847, p[5, 5] -> 0.00145385,
p[6, 2] -> 0.000261693, p[6, 3] -> 0.00039254, p[6, 4] -> 0.000436155,
p[7, 1] -> 0.000078508, p[7, 2] -> 0.000117762, p[7, 3] -> 0.000130847,
p[8, 0] -> 0.0000235524, p[8, 1] -> 0.0000353286, p[8, 2] -> 0.000039254,
p[9, 0] -> 0.0000176643, p[9, 1] -> 0.0000117762, p[10, 0] -> 8.83215*10^-6}
I want to extract a specific element from the list and add it.
For example, I want to extract the element which becomes 8 by the element of p [i, j] and add it.
p[0, 8] + p[1, 7] + p[2, 6] + ... = answer
Please advise an efficient way, thank you.
list-manipulation
list-manipulation
edited Jan 8 at 6:21
Αλέξανδρος Ζεγγ
4,46011029
asked Jan 8 at 5:37
Kenta KawaiKenta Kawai
385
edited Jan 8 at 6:21
Αλέξανδρος Ζεγγ
4,46011029
asked Jan 8 at 5:37
Kenta KawaiKenta Kawai
385
edited Jan 8 at 6:21
Αλέξανδρος Ζεγγ
4,46011029
edited Jan 8 at 6:21
Αλέξανδρος Ζεγγ
4,46011029
edited Jan 8 at 6:21
Αλέξανδρος Ζεγγ
4,46011029
4,46011029
asked Jan 8 at 5:37
Kenta KawaiKenta Kawai
385
asked Jan 8 at 5:37
Kenta KawaiKenta Kawai
385
asked Jan 8 at 5:37
Kenta KawaiKenta Kawai
385
385
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
7
$begingroup$
First assign a name to your above list, e.g., run
rules = (*your list*);
Then generate the expression you want, using Table or Array
sum = Plus @@ Table[p[i, 8 - i], {i, 0, 8}]
sum = Array[p[#, 8 - #] &, 9, 0, Plus]
p[0, 8] + p[1, 7] + p[2, 6] + p[3, 5] + p[4, 4] + p[5, 3] + p[6, 2] + p[7, 1] + p[8, 0]
Finally, apply the rules to finish the value substitution via ReplaceAll (/.):
sum /. rules
0.692301
answered Jan 8 at 6:26
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,46011029
$endgroup$
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
add a comment |
4
$begingroup$
list= (*your list*)
Total[Select[list, (#[[1, 1]] + #[[1, 2]]) == 8 &] [[All, 2]]]
(*0.692301*)
answered Jan 8 at 6:57
Hubble07Hubble07
2,990721
$endgroup$
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
$begingroup$
You can even useKeySelect:Total @ KeySelect[list, Plus @@ # == 8 &]
$endgroup$
– Sjoerd Smit
Jan 8 at 17:02
add a comment |
3
$begingroup$
Consider GroupBy or Merge. Pick is less specific but still applicable.
rules = {p[0, 8] -> 0.538464, p[1, 7] -> 0.107693, p[1, 8] -> 0.161539,
p[2, 6] -> 0.0323078, p[2, 7] -> 0.0484617, p[2, 8] -> 0.0538464,
p[3, 5] -> 0.00969234, p[3, 6] -> 0.0145385, p[3, 7] -> 0.0161539,
p[4, 4] -> 0.0029077, p[4, 5] -> 0.00436155, p[4, 6] -> 0.00484617,
p[5, 3] -> 0.000872311, p[5, 4] -> 0.00130847, p[5, 5] -> 0.00145385,
p[6, 2] -> 0.000261693, p[6, 3] -> 0.00039254, p[6, 4] -> 0.000436155,
p[7, 1] -> 0.000078508, p[7, 2] -> 0.000117762, p[7, 3] -> 0.000130847,
p[8, 0] -> 0.0000235524, p[8, 1] -> 0.0000353286, p[8, 2] -> 0.000039254,
p[9, 0] -> 0.0000176643, p[9, 1] -> 0.0000117762, p[10, 0] -> 8.83215*10^-6};
GroupBy[rules, Total@*First -> Last, Tr][8]
Merge[Total@# -> #2 & @@@ rules, Tr][8]
Merge[Plus @@@ # & /@ rules, Tr][8]
Merge[rules /. p -> Plus, Tr][8]
Pick[Values@rules, Total /@ Keys@rules, 8] // Tr
All give
0.692301
answered Jan 8 at 10:22
Mr.Wizard♦Mr.Wizard
232k294781063
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
$begingroup$
First assign a name to your above list, e.g., run
rules = (*your list*);
Then generate the expression you want, using Table or Array
sum = Plus @@ Table[p[i, 8 - i], {i, 0, 8}]
sum = Array[p[#, 8 - #] &, 9, 0, Plus]
p[0, 8] + p[1, 7] + p[2, 6] + p[3, 5] + p[4, 4] + p[5, 3] + p[6, 2] + p[7, 1] + p[8, 0]
Finally, apply the rules to finish the value substitution via ReplaceAll (/.):
sum /. rules
0.692301
answered Jan 8 at 6:26
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,46011029
$endgroup$
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
add a comment |
7
$begingroup$
First assign a name to your above list, e.g., run
rules = (*your list*);
Then generate the expression you want, using Table or Array
sum = Plus @@ Table[p[i, 8 - i], {i, 0, 8}]
sum = Array[p[#, 8 - #] &, 9, 0, Plus]
p[0, 8] + p[1, 7] + p[2, 6] + p[3, 5] + p[4, 4] + p[5, 3] + p[6, 2] + p[7, 1] + p[8, 0]
Finally, apply the rules to finish the value substitution via ReplaceAll (/.):
sum /. rules
0.692301
answered Jan 8 at 6:26
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,46011029
$endgroup$
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
add a comment |
7
7
7
$begingroup$
First assign a name to your above list, e.g., run
rules = (*your list*);
Then generate the expression you want, using Table or Array
sum = Plus @@ Table[p[i, 8 - i], {i, 0, 8}]
sum = Array[p[#, 8 - #] &, 9, 0, Plus]
p[0, 8] + p[1, 7] + p[2, 6] + p[3, 5] + p[4, 4] + p[5, 3] + p[6, 2] + p[7, 1] + p[8, 0]
Finally, apply the rules to finish the value substitution via ReplaceAll (/.):
sum /. rules
0.692301
answered Jan 8 at 6:26
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,46011029
$endgroup$
First assign a name to your above list, e.g., run
rules = (*your list*);
Then generate the expression you want, using Table or Array
sum = Plus @@ Table[p[i, 8 - i], {i, 0, 8}]
sum = Array[p[#, 8 - #] &, 9, 0, Plus]
p[0, 8] + p[1, 7] + p[2, 6] + p[3, 5] + p[4, 4] + p[5, 3] + p[6, 2] + p[7, 1] + p[8, 0]
Finally, apply the rules to finish the value substitution via ReplaceAll (/.):
sum /. rules
0.692301
answered Jan 8 at 6:26
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,46011029
answered Jan 8 at 6:26
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,46011029
answered Jan 8 at 6:26
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,46011029
answered Jan 8 at 6:26
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,46011029
4,46011029
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
add a comment |
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
add a comment |
4
$begingroup$
list= (*your list*)
Total[Select[list, (#[[1, 1]] + #[[1, 2]]) == 8 &] [[All, 2]]]
(*0.692301*)
answered Jan 8 at 6:57
Hubble07Hubble07
2,990721
$endgroup$
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
$begingroup$
You can even useKeySelect:Total @ KeySelect[list, Plus @@ # == 8 &]
$endgroup$
– Sjoerd Smit
Jan 8 at 17:02
add a comment |
4
$begingroup$
list= (*your list*)
Total[Select[list, (#[[1, 1]] + #[[1, 2]]) == 8 &] [[All, 2]]]
(*0.692301*)
answered Jan 8 at 6:57
Hubble07Hubble07
2,990721
$endgroup$
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
$begingroup$
You can even useKeySelect:Total @ KeySelect[list, Plus @@ # == 8 &]
$endgroup$
– Sjoerd Smit
Jan 8 at 17:02
add a comment |
4
4
4
$begingroup$
list= (*your list*)
Total[Select[list, (#[[1, 1]] + #[[1, 2]]) == 8 &] [[All, 2]]]
(*0.692301*)
answered Jan 8 at 6:57
Hubble07Hubble07
2,990721
$endgroup$
list= (*your list*)
Total[Select[list, (#[[1, 1]] + #[[1, 2]]) == 8 &] [[All, 2]]]
(*0.692301*)
answered Jan 8 at 6:57
Hubble07Hubble07
2,990721
answered Jan 8 at 6:57
Hubble07Hubble07
2,990721
answered Jan 8 at 6:57
Hubble07Hubble07
2,990721
answered Jan 8 at 6:57
Hubble07Hubble07
2,990721
2,990721
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
$begingroup$
You can even useKeySelect:Total @ KeySelect[list, Plus @@ # == 8 &]
$endgroup$
– Sjoerd Smit
Jan 8 at 17:02
add a comment |
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
$begingroup$
You can even useKeySelect:Total @ KeySelect[list, Plus @@ # == 8 &]
$endgroup$
– Sjoerd Smit
Jan 8 at 17:02
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
$begingroup$
Thank you for advice.It solved properly.
$endgroup$
– Kenta Kawai
Jan 8 at 7:53
$begingroup$
You can even use
KeySelect: Total @ KeySelect[list, Plus @@ # == 8 &]$endgroup$
– Sjoerd Smit
Jan 8 at 17:02
$begingroup$
You can even use
KeySelect: Total @ KeySelect[list, Plus @@ # == 8 &]$endgroup$
– Sjoerd Smit
Jan 8 at 17:02
add a comment |
3
$begingroup$
Consider GroupBy or Merge. Pick is less specific but still applicable.
rules = {p[0, 8] -> 0.538464, p[1, 7] -> 0.107693, p[1, 8] -> 0.161539,
p[2, 6] -> 0.0323078, p[2, 7] -> 0.0484617, p[2, 8] -> 0.0538464,
p[3, 5] -> 0.00969234, p[3, 6] -> 0.0145385, p[3, 7] -> 0.0161539,
p[4, 4] -> 0.0029077, p[4, 5] -> 0.00436155, p[4, 6] -> 0.00484617,
p[5, 3] -> 0.000872311, p[5, 4] -> 0.00130847, p[5, 5] -> 0.00145385,
p[6, 2] -> 0.000261693, p[6, 3] -> 0.00039254, p[6, 4] -> 0.000436155,
p[7, 1] -> 0.000078508, p[7, 2] -> 0.000117762, p[7, 3] -> 0.000130847,
p[8, 0] -> 0.0000235524, p[8, 1] -> 0.0000353286, p[8, 2] -> 0.000039254,
p[9, 0] -> 0.0000176643, p[9, 1] -> 0.0000117762, p[10, 0] -> 8.83215*10^-6};
GroupBy[rules, Total@*First -> Last, Tr][8]
Merge[Total@# -> #2 & @@@ rules, Tr][8]
Merge[Plus @@@ # & /@ rules, Tr][8]
Merge[rules /. p -> Plus, Tr][8]
Pick[Values@rules, Total /@ Keys@rules, 8] // Tr
All give
0.692301
answered Jan 8 at 10:22
Mr.Wizard♦Mr.Wizard
232k294781063
$endgroup$
add a comment |
3
$begingroup$
Consider GroupBy or Merge. Pick is less specific but still applicable.
rules = {p[0, 8] -> 0.538464, p[1, 7] -> 0.107693, p[1, 8] -> 0.161539,
p[2, 6] -> 0.0323078, p[2, 7] -> 0.0484617, p[2, 8] -> 0.0538464,
p[3, 5] -> 0.00969234, p[3, 6] -> 0.0145385, p[3, 7] -> 0.0161539,
p[4, 4] -> 0.0029077, p[4, 5] -> 0.00436155, p[4, 6] -> 0.00484617,
p[5, 3] -> 0.000872311, p[5, 4] -> 0.00130847, p[5, 5] -> 0.00145385,
p[6, 2] -> 0.000261693, p[6, 3] -> 0.00039254, p[6, 4] -> 0.000436155,
p[7, 1] -> 0.000078508, p[7, 2] -> 0.000117762, p[7, 3] -> 0.000130847,
p[8, 0] -> 0.0000235524, p[8, 1] -> 0.0000353286, p[8, 2] -> 0.000039254,
p[9, 0] -> 0.0000176643, p[9, 1] -> 0.0000117762, p[10, 0] -> 8.83215*10^-6};
GroupBy[rules, Total@*First -> Last, Tr][8]
Merge[Total@# -> #2 & @@@ rules, Tr][8]
Merge[Plus @@@ # & /@ rules, Tr][8]
Merge[rules /. p -> Plus, Tr][8]
Pick[Values@rules, Total /@ Keys@rules, 8] // Tr
All give
0.692301
answered Jan 8 at 10:22
Mr.Wizard♦Mr.Wizard
232k294781063
$endgroup$
add a comment |
3
3
3
$begingroup$
Consider GroupBy or Merge. Pick is less specific but still applicable.
rules = {p[0, 8] -> 0.538464, p[1, 7] -> 0.107693, p[1, 8] -> 0.161539,
p[2, 6] -> 0.0323078, p[2, 7] -> 0.0484617, p[2, 8] -> 0.0538464,
p[3, 5] -> 0.00969234, p[3, 6] -> 0.0145385, p[3, 7] -> 0.0161539,
p[4, 4] -> 0.0029077, p[4, 5] -> 0.00436155, p[4, 6] -> 0.00484617,
p[5, 3] -> 0.000872311, p[5, 4] -> 0.00130847, p[5, 5] -> 0.00145385,
p[6, 2] -> 0.000261693, p[6, 3] -> 0.00039254, p[6, 4] -> 0.000436155,
p[7, 1] -> 0.000078508, p[7, 2] -> 0.000117762, p[7, 3] -> 0.000130847,
p[8, 0] -> 0.0000235524, p[8, 1] -> 0.0000353286, p[8, 2] -> 0.000039254,
p[9, 0] -> 0.0000176643, p[9, 1] -> 0.0000117762, p[10, 0] -> 8.83215*10^-6};
GroupBy[rules, Total@*First -> Last, Tr][8]
Merge[Total@# -> #2 & @@@ rules, Tr][8]
Merge[Plus @@@ # & /@ rules, Tr][8]
Merge[rules /. p -> Plus, Tr][8]
Pick[Values@rules, Total /@ Keys@rules, 8] // Tr
All give
0.692301
answered Jan 8 at 10:22
Mr.Wizard♦Mr.Wizard
232k294781063
$endgroup$
Consider GroupBy or Merge. Pick is less specific but still applicable.
rules = {p[0, 8] -> 0.538464, p[1, 7] -> 0.107693, p[1, 8] -> 0.161539,
p[2, 6] -> 0.0323078, p[2, 7] -> 0.0484617, p[2, 8] -> 0.0538464,
p[3, 5] -> 0.00969234, p[3, 6] -> 0.0145385, p[3, 7] -> 0.0161539,
p[4, 4] -> 0.0029077, p[4, 5] -> 0.00436155, p[4, 6] -> 0.00484617,
p[5, 3] -> 0.000872311, p[5, 4] -> 0.00130847, p[5, 5] -> 0.00145385,
p[6, 2] -> 0.000261693, p[6, 3] -> 0.00039254, p[6, 4] -> 0.000436155,
p[7, 1] -> 0.000078508, p[7, 2] -> 0.000117762, p[7, 3] -> 0.000130847,
p[8, 0] -> 0.0000235524, p[8, 1] -> 0.0000353286, p[8, 2] -> 0.000039254,
p[9, 0] -> 0.0000176643, p[9, 1] -> 0.0000117762, p[10, 0] -> 8.83215*10^-6};
GroupBy[rules, Total@*First -> Last, Tr][8]
Merge[Total@# -> #2 & @@@ rules, Tr][8]
Merge[Plus @@@ # & /@ rules, Tr][8]
Merge[rules /. p -> Plus, Tr][8]
Pick[Values@rules, Total /@ Keys@rules, 8] // Tr
All give
0.692301
answered Jan 8 at 10:22
Mr.Wizard♦Mr.Wizard
232k294781063
edited Jan 8 at 10:52
answered Jan 8 at 10:22
Mr.Wizard♦Mr.Wizard
232k294781063
answered Jan 8 at 10:22
Mr.Wizard♦Mr.Wizard
232k294781063
answered Jan 8 at 10:22
Mr.Wizard♦Mr.Wizard
232k294781063
232k294781063
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