在python中评估嵌套字典

如何解决在python中评估嵌套字典

我有一本字典，看起来像这样

//@version=4
LOOKBACK = 1000
study("FRACTAL_LINES",shorttitle="FL2",overlay=true,precision=0,max_bars_back=LOOKBACK)
//////////////////////FRACTALS/////////////////////////////////
header_fractals         = input(false,title = "====== Fractal Settings ======")
display_fractals        = input(false,title="Display Fractal triangles")
fractal_join_line       = input(true,title='Display Fractal lines')
extend                  = input(true,title='Extend fractal lines to current bar if they remain uncrossed')

aggressive = false
price = hl2

// fractal calculation
n = 2
header_fractals4 = input(false,title = "========================")

// Identify FRACTAL TOPS
isBWFractalBullish(mode) => ret = mode == 1 ? ((high[n+2] < high[n]) and (high[n+1] < high[n]) and (high[n-1] < high[n]) 
 and (high[n-2] < high[n])) or ((high[n+3] < high[n]) and (high[n+2] < high[n]) 
 and (high[n+1] == high[n]) and (high[n-1] < high[n]) and (high[n-2] < high[n])) 
 or ((high[n+4] < high[n]) and (high[n+3] < high[n]) and (high[n+2] == high[n]) 
 and (high[n+1] <= high[n]) and (high[n-1] < high[n]) and (high[n-2] < high[n])) 
 or ((high[n+5] < high[n]) and (high[n+4] < high[n]) and (high[n+3] == high[n]) 
 and (high[n+2] == high[n]) and (high[n+1] <= high[n]) and (high[n-1] < high[n]) and (high[n-2] < high[n])) 
 or ((high[n+6] < high[n]) and (high[n+5] < high[n]) and (high[n+4] == high[n]) 
 and (high[n+3] <= high[n]) and (high[n+2] == high[n]) and (high[n+1] <= high[n]) 
 and (high[n-1] < high[n]) and (high[n-2] < high[n])) : false
 
// Identify FRACTAL BOTTOMS
isBWFractalBearish(mode) => ret = mode == -1 ? ((low[n+2] > low[n]) and (low[n+1] > low[n]) and (low[n-1] > low[n]) 
 and (low[n-2] > low[n]))or ((low[n+3] > low[n]) and (low[n+2] > low[n]) and (low[n+1] == low[n]) 
 and (low[n-1] > low[n]) and (low[n-2] > low[n])) or ((low[n+4] > low[n]) and (low[n+3] > low[n]) 
 and (low[n+2] == low[n]) and (low[n+1] >= low[n]) and (low[n-1] > low[n]) and (low[n-2] > low[n])) 
 or ((low[n+5] > low[n]) and (low[n+4] > low[n]) and (low[n+3] == low[n]) and (low[n+2] == low[n]) 
 and (low[n+1] >= low[n]) and (low[n-1] > low[n]) and (low[n-2] > low[n])) or ((low[n+6] > low[n]) 
 and (low[n+5] > low[n]) and (low[n+4] == low[n]) and (low[n+3] >= low[n]) and (low[n+2] == low[n]) 
 and (low[n+1] >= low[n]) and (low[n-1] > low[n]) and (low[n-2] > low[n])) : false
 
filteredtopf = isBWFractalBullish(1)
filteredbotf = isBWFractalBearish(-1)

plotshape(filteredtopf and display_fractals,title="Up-Fractal",style=shape.triangleup,location=location.abovebar,offset=-2,color=color.green,transp=0)
plotshape(filteredbotf and display_fractals,title="Down-Fractal",style=shape.triangledown,location=location.belowbar,color=color.red,transp=0)

//// FRACTAL TOPS AND BOTTOMS Plots //////
plot(fractal_join_line ? valuewhen(filteredtopf,high[2],0) : na,title="Fractal Tops",style=plot.style_cross,linewidth=1,color=#b71c1c,transp=0) 
plot(fractal_join_line ? valuewhen(filteredbotf,low[2],title="Fractal Bottoms",color=#1b5e20,transp=0) 


//// Draw fractal lines if extend option is TRUE
line fractal_top_line    = na
line fractal_bottom_line = na
int fractal_top_bar      = na
int fractal_bottom_bar   = na
yTop = valuewhen(filteredtopf,0)
yBot = valuewhen(filteredbotf,0)
if extend

    if filteredtopf
        fractal_top_line := line.new(x1=bar_index[2],y1=yTop,x2=bar_index,y2=yTop,color=color.red)
        line.set_extend(fractal_top_line,extend.right)
    if filteredbotf
        fractal_bottom_line := line.new(x1=bar_index[2],y1=yBot,y2=yBot,color=color.green)
        line.set_extend(fractal_bottom_line,extend.right)

    // Delete fractal lines if price crosses them
    // float linePrice = na
    // lineCrossed = false
    // for i=0 to LOOKBACK
    //     if not na(fractal_top_bar[i])
    //         linePrice := line.get_price(fractal_top_line[i],bar_index)
    //         lineCrossed := (close[i+1] < linePrice and close[i] > linePrice) or (close[i+1] > linePrice and close[i] < linePrice)
    //         if lineCrossed
    //             line.delete(fractal_top_line[i])
        
    //     if not na(fractal_bottom_bar[i])
    //         linePrice := line.get_price(fractal_bottom_line[i],bar_index)
    //         lineCrossed := (close[i+1] < linePrice and close[i] > linePrice) or (close[i+1] > linePrice and close[i] < linePrice)
    //         if lineCrossed
    //             line.delete(fractal_bottom_line[i]) 

我试图从中提取数据，但是我不知道这样做的最佳方法。

result = {
   {
   "Dogs":{
      "original":[
         72,15.34386
      ],"result":[
         1,4
      ]
   },"Cats":{
      "original":[
         24.716667,21.563121
      ],2,"Lions":{
      "original":[
         15.761111,15.761111,15.761111
      ],3,"Bulls":{
      "original":[
         7.961111,7.055556,44.45,35.644444,17.805556,70.988889,35.527778
      ],"ivupstream-lix-frontend-0/invisualize LATENCY campaign-manager-web to RESTLI lixFrontendTreatmentsV2 on lix-frontend":{
      "original":[
         37.386364,28.289394
      ],"isnake":{
      "original":[
         18.399561,18.399561,20.361258,18.674232,18.248904
      ],4
      ]
   }
}

尽管这样，我还是尝试了几种方法。

我想遍历字典并为每组字典键获取>>> for i in result.items(): ... print(i['Dogs']) ... Traceback (most recent call last): File "<stdin>",line 2,in <module> TypeError: tuple indices must be integers or slices,not str 和result。这样做的最Python方式是什么？

解决方法

.items在每次迭代中返回两个arg，键和值。

for k,v in result.items():
    print(k,v)

,

Zaven的答案更加冗长，更接近问题的要求：

for k,v in result.items():
   r = v["result"]
   o = v["original"]
   print(k,r,o)