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想问一下大家,李建平老师的矢量合成分析程序里要怎么读写数据呢,主要是根据指数来确定显著区域。很纠结,子程序如下:
!c-----*----------------------------------------------------6---------7--
!c For a time ve!!ctor series f(n,2), !!computing:
!c (1) mean ve!ctor in the high index years of x(n)
!c (2) mean ve!ctor in the low index years of x(n)
!c (3) !composite differen!ce between the mean ve!ctor of f(n,2) in high
!c index years of x(n) and its !climatology average
!c (4) !composite differen!ce between the mean ve!ctor of f(n,2) in low
!c index years of x(n) and its !climate average
!c (5) !composite differen!ce between the mean ve!ctors of f(n,2) in high
!c and low index years of x(n)
!c input: n,x(n),f(n,2),!coefh,!coefl
!c n: the length of time series
!c x: !control variable (index)
!c f: given series
!c !coefh: !control parameter for high index (i.e., high index years are
!c those in whi!ch x(i) > !coefh)
!c !coefl: !control parameter for low index (i.e., low index years are
!c those in whi!ch x(i) < !coefl)
!c output: fh(2),fl(2),dh(2),dl(2),dhl(2),tn(5)
!c fh: the mean ve!ctor of f in high index years of x(n)
!c fl: the mean ve!ctor of f in low index years of x(n)
!c dh: !composite differen!ce between the mean ve!ctor of f in high index years of x(n)
!c and its !climate mean (i.e., high index years minus !cliamte mean).
!c dl: !composite differen!ce between the mean ve!ctor of f in low index years of x(n)
!c and its !climate mean (i.e., low index years minus !cliamte mean).
!c dhl: !composite differen!ce between the mean ve!ctors of f in high and low index years
!c of x(n) (i.e., high minus low index years)
!c tn(i,j): tn only equals 1. or 0. !corresponding to signifi!cant differen!ce or not.
!c tn=1. indi!cates that the differen!ce is signifi!cant.
!c tn=0. indi!cates that the differen!ce is not signifi!cant.
!c tn(1,j)~tn(5,j) are !corresponding to the 90%,95%,98%,99% and 99.9% !confident levels.
!c j=1: siginifi!cant test for dh
!c j=2: siginifi!cant test for dl
!c j=3: siginifi!cant test for dhl
!c Mar!ch 10, 2004 by Jianping Li.
subroutine differhl1V(n,x,f,coefh,coefl,fh,fl,dh,dl,dhl,tn)
dimension x(n),f(n,2)
dimension fh(2),fl(2),dh(2),dl(2),dhl(2),tn(5,3)
real avef(2),hn(n,2),ln(n,2) !work array
do j=1,2
call meanvar(n,f(1,j),avef(j),sf,vf)
enddo
do j=1,2
nh=0
nl=0
do k=1,n
if(x(k).gt.coefh)then
nh=nh+1
hn(nh,j)=f(k,j)
endif
if(x(k).lt.coefl)then
nl=nl+1
ln(nl,j)=f(k,j)
endif
enddo
call meanvar1(nh,hn(1,j),fh(j))
call meanvar1(nl,ln(1,j),fl(j))
dh(j)=fh(j)-avef(j)
dl(j)=fl(j)-avef(j)
dhl(j)=fh(j)-fl(j)
enddo
call diff_t_testV1(n,nh,f(1,1),f(1,2),hn(1,1),hn(1,2),tn(1,1))
call diff_t_testV1(n,nl,f(1,1),f(1,2),ln(1,1),ln(1,2),tn(1,2))
call diff_t_testV1(nh,nl,hn(1,1),hn(1,2),ln(1,1),ln(1,2),tn(1,3))
return
end
!c-----*----------------------------------------------------6---------7--
!c Student's t-test for the differen!ce between two ve!ctor means
!c input: n,m,x(n,2),y(m,2),nt
!c n: number of x
!c m: number of y
!c x(n,2): ve!ctor series
!c y(m,2): ve!ctor series
!c nt: signifi!cant level, nt must be the one of 90,95,98,99 and 999
!c output: tn(5)
!c tn: tn only equals 1. or 0. !corresponding to signifi!cant differen!ce or not.
!c tn(1)~tn(5) are !corresponding to the 90%,95%,98%,99% and 99.9% !confident levels.
!c By Dr. Jianping Li, Mar!ch 05, 2004
subroutine diff_t_testV1(n,m,x1,x2,y1,y2,tn)
parameter(nn=10000)
real x(n,2),y(m,2),tn(5),x1(n),x2(n),y1(m),y2(m)
real ft(nn,5),ax(2),ay(2),wx(n,2),wy(m,2),wx1(n),wy1(m) !work array
do i=1,n
x(i,1)=x1(i)
x(i,2)=x2(i)
enddo
do i=1,m
y(i,1)=y1(i)
y(i,2)=y2(i)
enddo
do i=1,5
tn(i)=0.
enddo
do j=1,2
call meanvar1(n,x(1,j),ax(j))
call meanvar1(m,y(1,j),ay(j))
enddo
do j=1,2
do i=1,n
wx(i,j)=x(i,j)-ax(j)
enddo
do i=1,m
wy(i,j)=y(i,j)-ay(j)
enddo
enddo
do i=1,n
wx1(i)=sqrt(wx(i,1)**2+wx(i,2)**2)
enddo
do i=1,m
wy1(i)=sqrt(wy(i,1)**2+wy(i,2)**2)
enddo
call meanvar(n,wx1,ax1,sx1,vx1)
call meanvar(m,wy1,ay1,sy1,vy1)
sxy=(n*vx1+m*vy1)/(n+m-2.)
sxy=sxy*(1./n+1./m)
sxy=sqrt(sxy)
if(sxy.eq.0.)return
d1=ax(1)-ay(1)
d2=ax(2)-ay(2)
dxy=sqrt(d1**2+d2**2)
ta=dxy/sxy
nm=n+m-2
call t_table(ft(1,1),ft(1,2),ft(1,3),ft(1,4),ft(1,5))
do i=1,5
if(ta.ge.ft(nm,i))then
tn(i)=1.
endif
enddo
return
end
!c-----*----------------------------------------------------6---------7--
!c !computing the mean ax, standard deviation sx
!c and varian!ce vx of a series x(i) (i=1,...,n).
!c input: n and x(n)
!c n: number of raw series
!c x(n): raw series
!c output: ax, sx and vx
!c ax: the mean value of x(n)
!c sx: the standard deviation of x(n)
!c vx: the varian!ce of x(n)
!c By Dr. LI Jianping, May 6, 1998.
subroutine meanvar(n,x,ax,sx,vx)
dimension x(n)
ax=0.
vx=0.
sx=0.
do i=1,n
ax=ax+x(i)
enddo
ax=ax/float(n)
do i=1,n
vx=vx+(x(i)-ax)**2
enddo
vx=vx/float(n)
sx=sqrt(vx)
return
end
!c-----*----------------------------------------------------6---------7--
!c !computing the mean ax of a series x(i) (i=1,...,n).
!c input: n and x(n)
!c n: number of raw series
!c x(n): raw series
!c output: ax and sx
!c ax: the mean value of x(n)
!c By Dr. LI Jianping, May 6, 1999.
subroutine meanvar1(n,x,ax)
dimension x(n)
ax=0.
vx=0.
sx=0.
do i=1,n
ax=ax+x(i)
enddo
ax=ax/float(n)
return
end
!c-----*----------------------------------------------------6---------7--
!c t-distribution, i.e., student's distribution
!c t table with two-tailed (right and left tails) probabilities
!c P(|t|>=ta)=a
!c where a (alpha) signifi!can!ce level and (1-a)*100% is !confinden!ce level.
!c t90: t-distribution test at 90% !confiden!ce level.
!c t95: t-distribution test at 95% !confiden!ce level.
!c t98: t-distribution test at 98% !confiden!ce level.
!c t99: t-distribution test at 99% !confiden!ce level.
!c t999: t-distribution test at 99.9% !confiden!ce level.
!c By Dr. Jianping Li, January 5, 2000.
subroutine t_table(ft90,ft95,ft98,ft99,ft999)
parameter(n=10000)
dimension ft90(n),ft95(n),ft98(n),ft99(n),ft999(n)
dimension n90(30),n95(30),n98(30),n99(30),n999(30)
data n90/ 6314,2920,2353,2132,2015,1943,1895,1860,1833,1812,&
& 1796,1782,1771,1761,1753,1746,1740,1734,1729,1725,&
& 1721,1717,1714,1711,1708,1706,1703,1701,1699,1697/
data n95/12706,4303,3182,2776,2571,2447,2365,2306,2262,2228,&
& 2201,2179,2160,2145,2131,2120,2110,2101,2093,2086,&
& 2080,2074,2069,2064,2060,2056,2052,2048,2045,2042/
data n98/31821,6965,4541,3747,3365,3143,2998,2896,2821,2764,&
& 2718,2681,2650,2624,2602,2583,2567,2552,2539,2528,&
& 2518,2508,2500,2492,2485,2479,2473,2467,2462,2457/
data n99/63657,9925,5841,4604,4032,3707,3499,3355,3250,3169, &
& 3106,3055,3012,2977,2947,2921,2898,2878,2861,2845, &
& 2831,2819,2807,2797,2787,2779,2771,2763,2756,2750/
data n999/636619,31598,12941,8610,6859,5959,5405,5041,4781, &
& 4587,4437,4318,4221,4140,4073,4015,3965,3922,3883,3850, &
& 3819,3792,3767,3745,3725,3707,3690,3674,3659,3646/
do i=1,30
ft90(i)=n90(i)/1000.
ft95(i)=n95(i)/1000.
ft98(i)=n98(i)/1000.
ft99(i)=n99(i)/1000.
ft999(i)=n999(i)/1000.
enddo
do i=31,40
fi=(i-30.)/10.
ft90(i)=1.697-(1.697-1.684)*fi
ft95(i)=2.042-(2.042-2.021)*fi
ft98(i)=2.457-(2.457-2.423)*fi
ft99(i)=2.750-(2.750-2.704)*fi
ft999(i)=3.646-(3.646-3.551)*fi
enddo
do i=41,60
fi=(i-40.)/20.
ft90(i)=1.684-(1.684-1.671)*fi
ft95(i)=2.021-(2.021-2.000)*fi
ft98(i)=2.423-(2.423-2.390)*fi
ft99(i)=2.704-(2.704-2.660)*fi
ft999(i)=3.551-(3.551-3.460)*fi
enddo
do i=61,120
fi=(i-60.)/60.
ft90(i)=1.671-(1.671-1.658)*fi
ft95(i)=2.000-(2.000-1.980)*fi
ft98(i)=2.390-(2.390-2.358)*fi
ft99(i)=2.660-(2.660-2.617)*fi
ft999(i)=3.460-(3.460-3.373)*fi
enddo
do i=121,n
fi=(i-120.)/(n-120.)
ft90(i)=1.658-(1.658-1.645)*fi
ft95(i)=1.980-(1.980-1.960)*fi
ft98(i)=2.358-(2.358-2.326)*fi
ft99(i)=2.617-(2.617-2.576)*fi
ft999(i)=3.373-(3.373-2.291)*fi
enddo
return
end
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