Question

我需要的是如何过滤NSDictionary并仅返回键包含字符串的值和键

例如，如果我们NSDictionary包含：

{
  "houssam" : 3,
  "houss" : 2,
  "other" : 5
}

，字符串为"houss"

所以我们需要返回

{
  "houssam" : 3,
  "houss" : 2
}

最诚挚的问候，

Answer 1

您可以使用filter功能以下列方式获取所需内容：

var dict: NSDictionary  = ["houssam": 3, "houss": 2, "other": 5 ]
let string = "houss"

var result = dict.filter { $0.0.containsString(string)}
print(result) //[("houssam", 3), ("houss", 2)]

上面的代码返回一个元组列表，如果你想再次获得一个[String: Int]字典，你可以使用以下代码：

var newData = [String: Int]()
for x in result {
   newData[x.0 as! String] = x.1 as? Int
}

print(newData) //["houssam": 3, "houss": 2]

我希望这对你有所帮助。

Answer 2

使用此代码获取匹配的密钥。

var predicate = NSPredicate(format: "SELF like %@", "houss");
let matchingKeys = dictionary.keys.filter { predicate.evaluateWithObject($0) };

然后只需获取哪些键位于matchingKeys数组中。

Answer 3

由于这是一个NSDictionary，您可以在键数组上使用let data: NSDictionary = // Your NSDictionary let keys: NSArray = data.allKeys let filteredKeys: [String] = keys.filteredArrayUsingPredicate(NSPredicate(format: "SELF CONTAINS[cd] %@", "houss")) as! [String] let filteredDictionary = data.dictionaryWithValuesForKeys(filteredKeys)方法，并从初始字典中获取值。

例如：

from tkinter import *

#DATA
class Staff(object):
    def __init__(self, name, ID):
        self.name = name #this data comes from storage
        self.ID = ID #this is for this instance, starting from 0 (for use with grid)

ID42 = Staff("Joe", 0)
ID25 = Staff("George", 1)
ID84 = Staff("Eva", 2)

stafflist = [ID42, ID25, ID84]
weekdays = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"]

scheduleDictionary = {}

for r in range(0,3):
    scheduleDictionary[r] = ['shift','shift','shift','shift','shift','shift','shift','shift','shift','shift','shift','shift','shift','shift']

#Build window
root = Tk()

ScheduleFrame = Frame(root)
ScheduleFrame.pack()

#(Re)Build schedule on screen
def BuildSchedule():

    for r in range(1,4):
        Label(ScheduleFrame, text=stafflist[r-1].name).grid(row=r, column=0)

    for c in range(1,15):
        Label(ScheduleFrame, text=weekdays[(c-1)%7]).grid(row=0, column=c)

    for r in range(1,4):
        for c in range(1,15):
            Label(ScheduleFrame, text=scheduleDictionary[r-1][c-1]).grid(row=r, column=c)

#Mouse events
def mouse(event):
    y = event.widget.grid_info()['row'] - 1
    x = event.widget.grid_info()['column'] - 1
    print(x,y)
    shiftSelection(y,x)

#shiftSelection
def shiftSelection(row, column):
    shifts_window = Tk()
    box = Listbox(shifts_window)
    box.insert(1, "MR")
    box.insert(2, "AR")
    box.insert(3, "ER")
    box.pack()
    button = Button(shifts_window, text="Okay", command = lambda: selectShift(shifts_window, box.get(ACTIVE),row, column))
    button.pack()

def selectShift(shifts_window, shift,row, column):
    scheduleDictionary[row][column] = shift
    BuildSchedule()
    shifts_window.destroy()

root.bind("<Button-1>", mouse)

BuildSchedule()

root.mainloop()

希望有所帮助。

swift NSDictionary过滤器使用字符串

3 个答案: