表单提交反应后清除单选按钮

时间:2019-01-20 11:36:56

标签: react-native

{% for obj in your_qset %}
   {{ obj }}
{% endfor %}

我有一个带有单选按钮的简单表格 我想在提交后清除表格 但这并没有清除,这很奇怪

代码如下:

react-native-simple-radio-button

这是小吃的代码,您可以检查一下 https://snack.expo.io/@andreh111/aW5zYW

我在按下按钮时将initialRadioPos设置为-1,但是什么也没有发生,但是我认为代码逻辑是正确的……所以问题出在哪里

1 个答案:

答案 0 :(得分:1)

问题在于您尚未更改setwd("/home/greg/Documents/rwork/incid/bayes") library(bsts) library(mcmc) #simulate some data y<-rep(NA,50) y[1]=1 y[2]=1 s=2 set.seed(5) for (k in 1:48) { y[2+k]=y[1+k]+0.1*y[k]+s*rnorm(1) } plot(1:50,y[1:50],main=paste("seed =",5)) #bsts model ss<-AddLocalLevel(list(),y) mod1<-bsts(y,state.specification=ss,niter=1000) plot(mod1) #what is actually being plotted? #reproduce the plot using mod1$state.contributions par(mfrow=c(1,2)) plot(1:50,y,col="blue",ylim=c(-12,8),main="quantiles of mod1$state.contributions") for (i in 1:99) { qi<-qin<-rep(NA,50) tj<-1:50 for (j in 1:50) { qi[j]<-quantile(mod1$state.con[501:1000,1,j],(i-0.5)/100) qin[j]<-quantile(mod1$state.con[501:1000,1,j],(i+1-0.5)/100) } polygon(c(tj,rev(tj)),c(qi[1:50],rev(qin[1:50])),col=rgb(0,0,0,40*dnorm(i,mean=50,sd=20)),border=FALSE) } plot(mod1,ylim=c(-12,8),main="plot(mod1)") par(mfrow=c(1,1)) #the state specification is based on sd(y) sd(y) ss #kalman iteration, then add P error term when generating state from xe #set P[1,,] = 1 to start #then at end, reset P[1,,]=P[2,,] #polygon shading ax<-(mod1$sigma.level[501:1000])^2 bx<-(mod1$sigma.obs[501:1000])^2 state<-matrix(NA,500,50) for (j in 1:500) { a<-ax[j] b<-bx[j] H=matrix(1,1,1) F=matrix(1,1,1) N=50 dim(y)=c(1,N) xe<-ye<-ze<-matrix(NA,1,N) xe[,1]<-1 ye[,1]<-H%*%xe[,1] ze[,1]<-y[,1]-ye[,1] P<-K<-array(data=NA,dim=c(N,1,1)) P[1,,]<-1 for (i in 1:(N-1)) { P[i+1,,]<-F%*%P[i,,]%*%t(F)+a K[i+1,,]<-P[i+1,,]%*%t(H)%*%solve(b+H%*%P[i+1,,]%*%t(H)) P[i+1,,]<-(diag(1,1)-K[i+1,,]%*%H)%*%P[i+1,,] xe[1,i+1]<-F%*%xe[,i]+K[i+1,,]%*%(y[,i+1]-H%*%F%*%xe[,i]) } P[1,,]<-P[2,,] state[j,]<-rnorm(N,xe[1,1:N],P[1:N,,]^0.5) } par(mfrow=c(1,2)) plot(1:N,y,col="blue",ylim=c(-12,8),main="quantiles of kalman state") for (i in 1:99) { qi<-qin<-rep(NA,N) tj<-1:N for (k in 1:N) { qi[k]<-quantile(state[,k],(i-0.5)/100) qin[k]<-quantile(state[,k],(i+1-0.5)/100) } polygon(c(tj,rev(tj)),c(qi[1:N],rev(qin[1:N])),col=rgb(0,0,0,40*dnorm(i,mean=50,sd=20)),border=FALSE) } plot(mod1,ylim=c(-12,8)) par(mfrow=c(1,1)) #indices chosen at random #plausible that state[,i] and mod1$state.con[501:1000,1,i] are the same distribution par(mfrow=c(2,4),mar=c(5,5,3,1)) sami<-sample(1:50,8,replace=F) for (i in 1:8) { qqplot(state[,sami[i]],mod1$state.con[501:1000,1,sami[i]],cex=0.5,main=paste("i =",sami[i]),cex.lab=1.3,xlab=paste("state[,",sami[i],"]",sep=""),ylab=paste("mod1$state.con[501:1000,1,",sami[i],"]",sep="")) lines(state[,sami[i]],state[,sami[i]],col="red") } par(mfrow=c(1,1)) #the likelihood, using the kalman filter, as a function of the error variances and the initial state kal<-function(par) { a<-par[1] b<-par[2] init<-par[3] initP<-par[4] H=matrix(1,1,1) F=matrix(1,1,1) #1-dimensional state N=50 dim(y)=c(1,N) xe<-ye<-matrix(NA,1,N) xe[,1]<-init ye[,1]<-H%*%xe[,1] P<-K<-array(data=NA,dim=c(N,1,1)) #P[1,,] initial guess P[1,,]<-initP for (i in 1:(N-1)) { P[i+1,,]<-F%*%P[i,,]%*%t(F)+a K[i+1,,]<-P[i+1,,]%*%t(H)%*%solve(b+H%*%P[i+1,,]%*%t(H)) xe[1,i+1]<-F%*%xe[,i]+K[i+1,,]%*%(y[,i+1]-H%*%F%*%xe[,i]) P[i+1,,]<-(diag(1,1)-K[i+1,,]%*%H)%*%P[i+1,,] } -1/2*(log(abs(b))+(1/b)*sum((y[1,]-xe[1,])^2)) } #independent priors on a and b #bsts uses sd(y) to set both priors. #for a (ss[[1]]$sigma.prior) it uses SdPrior with #$prior.guess 0.04600655, $prior.df 0.01, $initial.value 0.04600655, $upper.limit 4.600655 #for b (mod1$prior) it uses SdPrior with #$prior.guess 4.600655, $prior.df 0.01, $initial.value 4.600655, $upper.limit 5.520786 #ss[[1]]$initial.state.prior is normal with mu=1, sd=4.600655 #try an inverse gamma prior on a and b, normal prior on init, with initP fixed initP<-0.25 lpr1<-function(a) { v=0.01 ifelse((a<=0)|a>(sd(y))^2,-Inf,-(v/2+1)*log(a)-v*(sd(y)/100)^2/(2*a)) } lpr2<-function(b) { v=0.01 ifelse((b<=0)|b>(1.2*sd(y))^2,-Inf,-(v/2+1)*log(b)-v*(sd(y))^2/(2*b)) } lpr3<-function(c) { -1/(2*sd(y))*(c-1)^2 } lpost1<-function(par) { a<-par[1] b<-par[2] init<-par[3] lpr1(a)+lpr2(b)+lpr3(init)+kal(c(par,initP)) } par0<-c(sd(y)/100,sd(y),1) nb=5000 out<-metrop(lpost1,par0,nb,scale=0.5) sam<-out$batch dim(sam) print(sam[500:510,],12) par(mfrow=c(3,1)) plot(sam[(nb/2):nb,1],type="l") plot(sam[(nb/2):nb,2],type="l") plot(sam[(nb/2):nb,3],type="l") par(mfrow=c(1,1)) op<-optim(par0,lpost1,control=list(fnscale=-1)) out<-metrop(lpost1,op$par,nb,scale=1) sam<-out$batch dim(sam) print(sam[500:510,],12) par(mfrow=c(3,1)) plot(sam[(nb/2):nb,1],type="l") plot(sam[(nb/2):nb,2],type="l") plot(sam[(nb/2):nb,3],type="l") par(mfrow=c(1,1)) mpar<-apply(sam[(nb/2):nb,],2,mean) lpost1m<-function(par) lpost1(par+mpar) opm<-optim(par0,lpost1m,control=list(fnscale=-1)) #a little fiddling gets plausible mixing nb=5000 #be patient outm<-metrop(lpost1m,opm$par,nb,blen=5,nspac=10,scale=1.5) outm$acc samm<-outm$batch dim(samm) par(mfrow=c(3,1)) plot(samm[(nb/2):nb,1],type="l") plot(samm[(nb/2):nb,2],type="l") plot(samm[(nb/2):nb,3],type="l") par(mfrow=c(1,1)) acf(samm) samm<-thin(samm,5) dim(samm) acf(samm) print(samm[501:510,],8) par(mfrow=c(3,1)) plot(samm[501:1000,1],type="l") plot(samm[501:1000,2],type="l") plot(samm[501:1000,3],type="l") par(mfrow=c(1,1)) #plot kalman using samm from lpost1m ax<-samm[501:1000,1]+mpar[1] bx<-samm[501:1000,2]+mpar[2] initx<-samm[501:1000,3]+mpar[3] state<-matrix(NA,500,50) for (j in 1:500) { a<-ax[j] b<-bx[j] init<-initx[j] H=matrix(1,1,1) F=matrix(1,1,1) N=50 dim(y)=c(1,N) xe<-ye<-matrix(NA,1,N) xe[,1]<-init ye[,1]<-H%*%xe[,1] P<-K<-array(data=NA,dim=c(N,1,1)) P[1,,]<-0 for (i in 1:(N-1)) { P[i+1,,]<-F%*%P[i,,]%*%t(F)+a K[i+1,,]<-P[i+1,,]%*%t(H)%*%solve(b+H%*%P[i+1,,]%*%t(H)) P[i+1,,]<-(diag(1,1)-K[i+1,,]%*%H)%*%P[i+1,,] xe[1,i+1]<-F%*%xe[,i]+K[i+1,,]%*%(y[,i+1]-H%*%F%*%xe[,i]) } P[1,,]<-P[2,,] state[j,]<-rnorm(N,xe[1,1:N],P[1:N,,]^0.5) } par(mfrow=c(1,2)) plot(1:N,y,col="blue",ylim=c(-12,8)) for (i in 1:99) { qi<-qin<-rep(NA,N) tj<-1:N for (k in 1:N) { qi[k]<-quantile(state[,k],(i-0.5)/100) qin[k]<-quantile(state[,k],(i+1-0.5)/100) } polygon(c(tj,rev(tj)),c(qi[1:N],rev(qin[1:N])),col=rgb(0,0,0,35*dnorm(i,mean=50,sd=15)),border=FALSE) } plot(mod1,ylim=c(-12,8)) par(mfrow=c(1,1)) #plausible #but mean(ax) mean((mod1$sigma.level[501:1000])^2) qqplot(ax,(mod1$sigma.level[501:1000])^2) mean(bx) mean((mod1$sigma.obs[501:1000])^2) qqplot(bx,(mod1$sigma.obs[501:1000])^2) lines(bx,bx,col="red") #so ax is not sampling (sigma.level)^2 #bx not sampling (sigma.obs)^2 的值,因此,在调用setState时,它将检查是否有任何更改,并注意是否相同,并且不会重新呈现组件。

您可以执行以下操作重新渲染组件。这有点 ,但是可以用。通过将RadioForm上的key属性设置为状态值,并在清除它时更新该值,它应该重新呈现该窗体。

代码如下:

initialRadioPos

https://snack.expo.io/HJXSxkGQV

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