Take single element of a list, and compare to average of whole list
If I have a list
lst = {1,2,3,4,5};
Is there someway of taking the first element and then comparing it to the average of the other 4? Then to take element 2 and compare it to the average of element 1,3,4,5? I can see how to do this with a For loop and a procedural style, but I am wondering if an easy functional style solution is possible.
Similar to This:
1 == Mean[{2,3,4,5}]
2 == Mean[{1,3,4,5}]
3 == Mean[{1,2,4,5}]
PS. I plan on using a more elaborate comparison, but == seemed like an easy one for this example.
list-manipulation functional-style averages
asked 2 hours ago
olliepower
1,21511127
add a comment |
If I have a list
lst = {1,2,3,4,5};
Is there someway of taking the first element and then comparing it to the average of the other 4? Then to take element 2 and compare it to the average of element 1,3,4,5? I can see how to do this with a For loop and a procedural style, but I am wondering if an easy functional style solution is possible.
Similar to This:
1 == Mean[{2,3,4,5}]
2 == Mean[{1,3,4,5}]
3 == Mean[{1,2,4,5}]
PS. I plan on using a more elaborate comparison, but == seemed like an easy one for this example.
list-manipulation functional-style averages
asked 2 hours ago
olliepower
1,21511127
add a comment |
If I have a list
lst = {1,2,3,4,5};
Is there someway of taking the first element and then comparing it to the average of the other 4? Then to take element 2 and compare it to the average of element 1,3,4,5? I can see how to do this with a For loop and a procedural style, but I am wondering if an easy functional style solution is possible.
Similar to This:
1 == Mean[{2,3,4,5}]
2 == Mean[{1,3,4,5}]
3 == Mean[{1,2,4,5}]
PS. I plan on using a more elaborate comparison, but == seemed like an easy one for this example.
list-manipulation functional-style averages
asked 2 hours ago
olliepower
1,21511127
If I have a list
lst = {1,2,3,4,5};
Is there someway of taking the first element and then comparing it to the average of the other 4? Then to take element 2 and compare it to the average of element 1,3,4,5? I can see how to do this with a For loop and a procedural style, but I am wondering if an easy functional style solution is possible.
Similar to This:
1 == Mean[{2,3,4,5}]
2 == Mean[{1,3,4,5}]
3 == Mean[{1,2,4,5}]
PS. I plan on using a more elaborate comparison, but == seemed like an easy one for this example.
list-manipulation functional-style averages
list-manipulation functional-style averages
asked 2 hours ago
olliepower
1,21511127
asked 2 hours ago
olliepower
1,21511127
asked 2 hours ago
olliepower
1,21511127
asked 2 hours ago
olliepower
1,21511127
asked 2 hours ago
olliepower
1,21511127
1,21511127
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
if you want to choose a specific element use
F[x_] := Mean@Complement[lst, {x}]
1 == F[1]
2 == F[2]
3 == F[3]
False
False
True
Note
if you want to choose the first,second,nth element use
F[x_] := Mean@Complement[lst, {lst[[x]]}]
1 == F[1]
2 == F[2]
3 == F[3]
False
False
True
answered 2 hours ago
J42161217
3,732220
add a comment |
ClearAll[subMeans]
subMeans = (Total[#] - #)/(Length[#] - 1) &;
Examples:
lst = {a, b, c, d};
subMeans[lst]
{1/3 (b + c + d), 1/3 (a + c + d), 1/3 (a + b + d), 1/3 (a + b + c)}
MapThread[Equal, {#, subMeans @ #}]& @ lst
{a == 1/3 (b + c + d), b == 1/3 (a + c + d), c == 1/3 (a + b + d),
d == 1/3 (a + b + c)}
MapThread[Equal, {#, subMeans @ #}]& @ Range[5]
{False, False, True, False, False}
answered 53 mins ago
kglr
177k9198405
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
if you want to choose a specific element use
F[x_] := Mean@Complement[lst, {x}]
1 == F[1]
2 == F[2]
3 == F[3]
False
False
True
Note
if you want to choose the first,second,nth element use
F[x_] := Mean@Complement[lst, {lst[[x]]}]
1 == F[1]
2 == F[2]
3 == F[3]
False
False
True
answered 2 hours ago
J42161217
3,732220
add a comment |
if you want to choose a specific element use
F[x_] := Mean@Complement[lst, {x}]
1 == F[1]
2 == F[2]
3 == F[3]
False
False
True
Note
if you want to choose the first,second,nth element use
F[x_] := Mean@Complement[lst, {lst[[x]]}]
1 == F[1]
2 == F[2]
3 == F[3]
False
False
True
answered 2 hours ago
J42161217
3,732220
add a comment |
if you want to choose a specific element use
F[x_] := Mean@Complement[lst, {x}]
1 == F[1]
2 == F[2]
3 == F[3]
False
False
True
Note
if you want to choose the first,second,nth element use
F[x_] := Mean@Complement[lst, {lst[[x]]}]
1 == F[1]
2 == F[2]
3 == F[3]
False
False
True
answered 2 hours ago
J42161217
3,732220
if you want to choose a specific element use
F[x_] := Mean@Complement[lst, {x}]
1 == F[1]
2 == F[2]
3 == F[3]
False
False
True
Note
if you want to choose the first,second,nth element use
F[x_] := Mean@Complement[lst, {lst[[x]]}]
1 == F[1]
2 == F[2]
3 == F[3]
False
False
True
answered 2 hours ago
J42161217
3,732220
edited 2 hours ago
answered 2 hours ago
J42161217
3,732220
answered 2 hours ago
J42161217
3,732220
answered 2 hours ago
J42161217
3,732220
3,732220
add a comment |
add a comment |
ClearAll[subMeans]
subMeans = (Total[#] - #)/(Length[#] - 1) &;
Examples:
lst = {a, b, c, d};
subMeans[lst]
{1/3 (b + c + d), 1/3 (a + c + d), 1/3 (a + b + d), 1/3 (a + b + c)}
MapThread[Equal, {#, subMeans @ #}]& @ lst
{a == 1/3 (b + c + d), b == 1/3 (a + c + d), c == 1/3 (a + b + d),
d == 1/3 (a + b + c)}
MapThread[Equal, {#, subMeans @ #}]& @ Range[5]
{False, False, True, False, False}
answered 53 mins ago
kglr
177k9198405
add a comment |
ClearAll[subMeans]
subMeans = (Total[#] - #)/(Length[#] - 1) &;
Examples:
lst = {a, b, c, d};
subMeans[lst]
{1/3 (b + c + d), 1/3 (a + c + d), 1/3 (a + b + d), 1/3 (a + b + c)}
MapThread[Equal, {#, subMeans @ #}]& @ lst
{a == 1/3 (b + c + d), b == 1/3 (a + c + d), c == 1/3 (a + b + d),
d == 1/3 (a + b + c)}
MapThread[Equal, {#, subMeans @ #}]& @ Range[5]
{False, False, True, False, False}
answered 53 mins ago
kglr
177k9198405
add a comment |
ClearAll[subMeans]
subMeans = (Total[#] - #)/(Length[#] - 1) &;
Examples:
lst = {a, b, c, d};
subMeans[lst]
{1/3 (b + c + d), 1/3 (a + c + d), 1/3 (a + b + d), 1/3 (a + b + c)}
MapThread[Equal, {#, subMeans @ #}]& @ lst
{a == 1/3 (b + c + d), b == 1/3 (a + c + d), c == 1/3 (a + b + d),
d == 1/3 (a + b + c)}
MapThread[Equal, {#, subMeans @ #}]& @ Range[5]
{False, False, True, False, False}
answered 53 mins ago
kglr
177k9198405
ClearAll[subMeans]
subMeans = (Total[#] - #)/(Length[#] - 1) &;
Examples:
lst = {a, b, c, d};
subMeans[lst]
{1/3 (b + c + d), 1/3 (a + c + d), 1/3 (a + b + d), 1/3 (a + b + c)}
MapThread[Equal, {#, subMeans @ #}]& @ lst
{a == 1/3 (b + c + d), b == 1/3 (a + c + d), c == 1/3 (a + b + d),
d == 1/3 (a + b + c)}
MapThread[Equal, {#, subMeans @ #}]& @ Range[5]
{False, False, True, False, False}
answered 53 mins ago
kglr
177k9198405
edited 46 mins ago
answered 53 mins ago
kglr
177k9198405
answered 53 mins ago
kglr
177k9198405
answered 53 mins ago
kglr
177k9198405
177k9198405
add a comment |
add a comment |
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