如何统一间隔?
How to unite intervals?
请帮助我 合并结果区间!
行“For...
”输出根存在的区间(根:2.94和5,52)。
我必须 考虑一下评论:
If in the intervals {x*[i],x[i+1]} and {x[i+1],x**[i+1]} can be the roots of the equation, the range {x*[i],x**[i+1]} must have at least one of its root.
X = {-2, 6}
spx = {-2, -1.90577, -1.81153, -1.59327, -1.375, -1.35785, -1.3407, -1.24655, -1.22941, -1.11811, -0.934054, -0.80167, -0.75, -0.625,-0.5, -0.25, -0.0981238, 0.303752, 0.651876, 0.94833, 1, 1.5, 1.75,2.11731, 2.5, 2.5625, 2.625, 3.3125, 3.75, 4, 4.00964, 4.01928,4.25964, 4.36731, 4.5, 4.75, 5, 5.25, 5.5, 5.75, 6}
spfw = {33.3632, 43.263, 51.6709, 55.5421, 57.1266, 57.2511, 57.3756,58.059, 58.0778, 58.1995,56.846,55.1903,54.5739,53.0828,51.1542,48.9959,48.0325,42.2533, 36.408,30.7952,30.1551,28.6446,23.138,19.4168,6.47053,5.90328,5.32951,-0.513959, -0.750527, -6.38895, -6.39157, -6.39418,-6.36456, -6.09357, -6.28599, -5.25369, -4.19539, -2.18625, -0.133803,2.90414, 6.171}
spfn = {33.3632, 40.2933, 46.5882, 51.9781, 55.5583, 55.5708, 55.5762, 55.4604, 55.4393, 55.0045, 530116, 51.1309, 50.4546, 48.1226, 45.6012, 43.402, 42.066, 37.5522, 32.6864, 28.1979, 28.0685,25.7067, 17.5943,13.5547, -2.97428, -3.21054, -3.36422, -5.05466, -5.1301, -6.4392,-6.76879, -6.48231, -7.20196, -7.00719, -7.53373, -6.00246, -4.41058,-2.8187, -1.16621, 2.35765, 6.04694}
For[i = 1, i < Length@spfn, i++,
If[! (((0 < spfn[[i]]) && (0 < spfn[[i + 1]])) ||
((spfw[[i]] < 0) && (spfw[[i + 1]] < 0))),
Print["1) exists root on: {", spx[[i]], ";", spx[[i + 1]], "}"]]]
所以结果包括 5 个区间:
1) exists root on: {2.11731;2.5}
1) exists root on: {2.5;2.5625}
1) exists root on: {2.5625;2.625}
1) exists root on: {2.625;3.3125}
1) exists root on: {5.5;5.75}
并且由于第一个根是 2.94,它必须进入 4 个第一个区间,最后一个是 5.52。所以在考虑了这句话之后,那条线应该输出两个间隔。
我尝试使用 IntervalUnion
但它不起作用:(
请帮我把这句话编码在这一行。
修改后的答案。有关 ## &[]
.
的说明,请参见 this link
spxpairs = Interval /@ Partition[spx, 2, 1];
spfwpairs = Interval /@ Partition[spfw, 2, 1];
spfnpairs = Interval /@ Partition[spfn, 2, 1];
ans = Apply[IntervalUnion,
If[Not[0 < Min@#1 || Max@#2 < 0], #3, ## &[]] &
@@@ Transpose[{spfnpairs, spfwpairs, spxpairs}]];
Column[Prepend[List @@ ans, "Roots exist on :"], Spacings -> 1]
Roots exist on :
{2.11731, 3.3125}
{5.5, 5.75}
是的,有一种更简洁的方法。一旦你有了间隔列表(如克里斯的回答),你就可以这样做:
IntervalUnion @@ intervals
(* 区间[{2.11731, 3.3125}, {5.5, 5.75}] *)
请帮助我 合并结果区间!
行“For...
”输出根存在的区间(根:2.94和5,52)。
我必须 考虑一下评论:
If in the intervals {x*[i],x[i+1]} and {x[i+1],x**[i+1]} can be the roots of the equation, the range {x*[i],x**[i+1]} must have at least one of its root.
X = {-2, 6}
spx = {-2, -1.90577, -1.81153, -1.59327, -1.375, -1.35785, -1.3407, -1.24655, -1.22941, -1.11811, -0.934054, -0.80167, -0.75, -0.625,-0.5, -0.25, -0.0981238, 0.303752, 0.651876, 0.94833, 1, 1.5, 1.75,2.11731, 2.5, 2.5625, 2.625, 3.3125, 3.75, 4, 4.00964, 4.01928,4.25964, 4.36731, 4.5, 4.75, 5, 5.25, 5.5, 5.75, 6}
spfw = {33.3632, 43.263, 51.6709, 55.5421, 57.1266, 57.2511, 57.3756,58.059, 58.0778, 58.1995,56.846,55.1903,54.5739,53.0828,51.1542,48.9959,48.0325,42.2533, 36.408,30.7952,30.1551,28.6446,23.138,19.4168,6.47053,5.90328,5.32951,-0.513959, -0.750527, -6.38895, -6.39157, -6.39418,-6.36456, -6.09357, -6.28599, -5.25369, -4.19539, -2.18625, -0.133803,2.90414, 6.171}
spfn = {33.3632, 40.2933, 46.5882, 51.9781, 55.5583, 55.5708, 55.5762, 55.4604, 55.4393, 55.0045, 530116, 51.1309, 50.4546, 48.1226, 45.6012, 43.402, 42.066, 37.5522, 32.6864, 28.1979, 28.0685,25.7067, 17.5943,13.5547, -2.97428, -3.21054, -3.36422, -5.05466, -5.1301, -6.4392,-6.76879, -6.48231, -7.20196, -7.00719, -7.53373, -6.00246, -4.41058,-2.8187, -1.16621, 2.35765, 6.04694}
For[i = 1, i < Length@spfn, i++,
If[! (((0 < spfn[[i]]) && (0 < spfn[[i + 1]])) ||
((spfw[[i]] < 0) && (spfw[[i + 1]] < 0))),
Print["1) exists root on: {", spx[[i]], ";", spx[[i + 1]], "}"]]]
所以结果包括 5 个区间:
1) exists root on: {2.11731;2.5}
1) exists root on: {2.5;2.5625}
1) exists root on: {2.5625;2.625}
1) exists root on: {2.625;3.3125}
1) exists root on: {5.5;5.75}
并且由于第一个根是 2.94,它必须进入 4 个第一个区间,最后一个是 5.52。所以在考虑了这句话之后,那条线应该输出两个间隔。
我尝试使用 IntervalUnion
但它不起作用:(
请帮我把这句话编码在这一行。
修改后的答案。有关 ## &[]
.
spxpairs = Interval /@ Partition[spx, 2, 1];
spfwpairs = Interval /@ Partition[spfw, 2, 1];
spfnpairs = Interval /@ Partition[spfn, 2, 1];
ans = Apply[IntervalUnion,
If[Not[0 < Min@#1 || Max@#2 < 0], #3, ## &[]] &
@@@ Transpose[{spfnpairs, spfwpairs, spxpairs}]];
Column[Prepend[List @@ ans, "Roots exist on :"], Spacings -> 1]
Roots exist on :
{2.11731, 3.3125}
{5.5, 5.75}
是的,有一种更简洁的方法。一旦你有了间隔列表(如克里斯的回答),你就可以这样做:
IntervalUnion @@ intervals
(* 区间[{2.11731, 3.3125}, {5.5, 5.75}] *)