Question

所以我有一个带有值的字典：[假设不重复]

require(corrplot)
#Reorder data collected in mtcars in this way: 
mtcars2 <- mtcars[ , c("carb", "wt", "hp", "cyl", "disp", 
                       "qsec", "vs",
                       "mpg", "drat", "am", "gear")
                   ]
#Let's say I have the following groups corresponding to these variables
groups <- c("","Engine",rep("", 3), "Speed","", "Fuel",rep("", 3) )
rows <- c("Engine","",rep("", 3), "Speed","", "Fuel",rep("", 3) )

R <- cor(mtcars2)
colnames(R) <- groups 
rownames(R) <- rows
corrplot(R, diag = FALSE, method = "ellipse", type = "upper", tl.col = "black")

我想做的就是使用值获取他们的键，所以如果给定1，我希望它获取具有值1（即0）并将其附加到列表中的键。

mydict={0: 1, 1: 2, -2: -1, -1: 0, -3: -2}

我不想创建for循环，因为我想在线性时间内完成。

Answer 1

如果可以确保您的值唯一，则可以创建dict的反向版本，在其中切换键和值：

mydict = {0: 1, 1: 2, -2: -1, -1: 0, -3: -2}

my_inverted_dict = dict(map(reversed, mydict.items()))

print(my_inverted_dict)
print(my_inverted_dict[1])

输出：

{1: 0, 2: 1, -1: -2, 0: -1, -2: -3}
0

Answer 2

您将为此循环，因为您必须反转字典的映射。为此，您可以使用字典理解：

Get-Content $alertlog|select-string $(Get-Date -format "ddd MMM dd")  -Context 0,1

Answer 3

mydict={0: 1, 1: 2, -2: -1, -1: 0, -3: -2}

search_value = 1

# List comprehension to return items if they match a search value
# This populates a list automatically
final0 = [k for k, v in mydict.items() if v == search_value]

# Note: If you have a value that is paired with more than one key
# then you might want to use set() and iterate through that to ensure
# your values aren't duplicates.

mydict2 = {0: 1, 1: 2, -2: -1, -1: 0, -3: -2, 4: 1} # Added matching value at end
final0 = [i for i in set([k for k, v in mydict.items() if v == search_value])]

Python：如何在不循环的情况下获取指定值的dict键？

3 个答案: