2017-01-09 37 views
-2
global_sulfur=0 
global_charcoal=0 
global_gun_powder=0 
global_low_grade_fuel=0 
global_metal_frags=0 
global_stone=0 
global_tech_trash=0 
global_cloth=0 
global_explosives=0 
global_rope=0 
global_small_stash=0 
global_beancan_grenade=0 
global_metal_pipe=0 
global_spring=0 
global_animal_fat=0 

def calculate(): 
    list={"Sulfur" : global_sulfur ,"Charcoal" : global_charcoal ,"Gun Powder" : global_gun_powder ,"Low Grade Fuel" : global_low_grade_fuel , "Metal Frags" : global_metal_frags , "Stone" : global_stone , \ 
     "Tech Trash" : global_tech_trash ,"Cloth" : global_cloth ,"Explosives" : global_explosives ,"Rope" : global_rope ,"Small Stash" : global_small_stash ,"Beancan Grenades" : global_beancan_grenade ,"Metal Pipe" : global_metal_pipe \ 
     ,"Spring" : global_spring ,"Animal Fat" : global_animal_fat} 

Theres 700+ more lines,but we want to make this questions simple,just think of the least of the variables will have a value over one。如果字典中的語句打印一個鍵和值,如果該值大於一,我該怎麼辦?

我試過多種方式和無數的時間,試圖找到一種方法,使該值是否超過鍵值,將打印價值和鍵超過1

+2

你不應該影響'list',特別是不能用*實際上不是列表*的東西。你也應該閱讀教程。 – jonrsharpe

回答

2

使用dict.items循環功能對。

data = { ... } 

for k, v in data.items(): 
    if v > 1: 
     print(k, v) 
+0

謝謝,就是這樣。如果你不介意,你可以解釋'K,V',我不明白它是如何工作的,除了它使兩個不同的變量,謝謝 – Seavan

+0

'K'和'V'是我給每個鍵和值的名稱爲對迭代通過。 –

相關問題