我有一个包含对象的数组。
这些对象具有fullname属性和member-id。
虽然fullname始终可用，但如果member-id是潜在会员但尚未注册，则member-id可以为null。
结果应如下所示：
{ id: 1122, name: Adrianna Budzinski }
{ id: 3785, name: Amy Divine }
{ id: 1555, name: Gale Purdue }
{ id: 1920, name: Rex Feng }
{ id: 2010, name: Samella Vizcaino }
{ id: null, name: Bethanie Weaver }
{ id: null, name: Celesta Gullo }
{ id: null, name: Darrick Fort }
{ id: null, name: Edmundo Boulanger }
{ id: null, name: Freddie Lanclos }
{ id: null, name: Gregory Lickteig }
{ id: null, name: Gwendolyn Cuadra }
{ id: null, name: Krystal Brosnahan }
{ id: null, name: Lahoma Pagani }
{ id: null, name: Senaida Risk }
{ id: null, name: Valarie Lopes }
具有id的成员应该按名称排在最前面，而不应该跟随的成员也按名称排序。
到目前为止我所取得的是两个单独的排序函数，但我不知道如何合并它们。
let sortedFriends = friends.sort(function(a, b){
if(a.name < b.name) return -1;
if(a.name > b.name) return 1;
return 0;
});
sortedFriends = sortedFriends.sort(function(a, b){
if(a.id === null) return 1;
if(b.id === null) return -1;
if(a.id === b.id) return 0;
if(a.id < b.id) return -1 ;
if(a.id < b.id) return 1;
});
答案 0 :(得分：2)
您可以按组（id或null）和名称进行排序。
var array = [{ id: null, name: 'Darrick Fort' }, { id: null, name: 'Edmundo Boulanger' }, { id: 1122, name: 'Adrianna Budzinski' }, { id: null, name: 'Freddie Lanclos' }, { id: null, name: 'Gregory Lickteig' }, { id: null, name: 'Gwendolyn Cuadra' }, { id: 2010, name: 'Samella Vizcaino' }, { id: null, name: 'Bethanie Weaver' }, { id: null, name: 'Celesta Gullo' }, { id: 3785, name: 'Amy Divine' }, { id: null, name: 'Krystal Brosnahan' }, { id: null, name: 'Lahoma Pagani' }, { id: 1920, name: 'Rex Feng' }, { id: 1555, name: 'Gale Purdue' }, { id: null, name: 'Senaida Risk' }, { id: null, name: 'Valarie Lopes' }];
array.sort(function (a, b) {
return (a.id === null) - (b.id === null) || a.name.localeCompare(b.name);
});
console.log(array);
答案 1 :(得分：1)
您可以将它们组合如下
import Tkinter,visa,time
from ttk import *
root = Tkinter.Tk
rm = visa.ResourceManager()
# Split port touple into a list
inst_list = []
str_inst = rm.list_resources()
for i in str_inst:
inst_list.append(i.split(","))
# Convert list into a string and replace it
def delChar(inst,input_inst):
str_new = ''.join(str(e) for e in inst_list[input_inst])
find_char = "()u[]"
for char in find_char:
inst = str_new.replace(char,"")
return inst
def delCharIn(list_del):
str_new = ''.join(str(e) for e in list_del)
find_char = "()u[]"
for char in find_char:
list_del = str_new.replace(char,"")
return list_del
# List the instruments and choose one
def instControl(inst):
new_list = []
for index in range(len(inst)):
new_list.append(delCharIn(inst[index]))
return new_list
#keithley = rm.open_resource(instControl(inst_list))
class InterfaceApp(root):
def __init__(self,parent):
root.__init__(self,parent)
self.parent = parent
self.initialize()
def initialize(self):
frInstrument = Tkinter.Frame(width=800, height=200, bg="",
colormap="new")
frInstrument.grid(row=0,sticky='EW')
separator = Tkinter.Frame(width=800, height=2, bd=1, relief='sunken')
separator.grid(row=1)
frSettings = Tkinter.Frame(width=800, height = 400, bg="",
colormap="new")
frSettings.grid(row=3,sticky='EW')
self.grid_columnconfigure(0,weight=1)
self.resizable(False,False)
groupInstrument = Tkinter.LabelFrame(frInstrument,
text="Choose an Instruments",
padx = 5, pady=5)
groupInstrument.grid(row = 0, column = 1, sticky="EW")
choices = instControl(inst_list)
self.choiceVarPower = Tkinter.StringVar()
self.choiceVarPower.set(choices[0])
inst1 =Combobox(groupInstrument, values = choices,textvariable =
self.choiceVarPower)
inst1.grid(row=0, column=2,sticky="EW",padx=2,pady=2)
instLabel1 = Tkinter.Label(groupInstrument, text="Power Supply: ")
instLabel1.grid(row=0,column=1,sticky='EW')
instLabel2 = Tkinter.Label(groupInstrument, text="Multimeter: ")
instLabel2.grid(row=1,column=1,sticky='EW')
self.choiceVarMulti = Tkinter.StringVar()
self.choiceVarMulti.set(choices[0])
inst2 = Combobox(groupInstrument, values=choices, textvariable =
self.choiceVarMulti)
inst2.grid(row=1,column=2,sticky='EW',padx=2,pady=2)
but_start = Tkinter.Button(frSettings,text='Run',
command=self.Run)
but_start.grid(row=0,column=1,sticky='EW')
but_closeInst = Tkinter.Button(frSettings,text="Close all instruments",
command = self.closeInst)
but_closeInst.grid(row=0,column=2,sticky='EW')
def W_KPower(scpi_comm):
return self.keithleyPower.write(":syst:loc")
def closeInst(self):
W_KPower(":syst:loc")
def Run(self):
self.keithleyPower = rm.open_resource(self.choiceVarPower.get())
self.keithleyMultimeter = rm.open_resource(self.choiceVarMulti.get())
if __name__ == '__main__':
app = InterfaceApp(None)
app.title("")
app.mainloop()
或使用 String#localeCompare() 方法
var friends = [{ id: 1122, name: 'Adrianna Budzinski'}, { id: 3785, name: 'Amy Divine'}, { id: 1555, name: 'Gale Purdue'}, { id: 1920, name: 'Rex Feng'}, { id: 2010, name: 'Samella Vizcaino'}, { id: null, name: 'Bethanie Weaver'}, { id: null, name: 'Celesta Gullo'}, { id: null, name: 'Darrick Fort'}, { id: null, name: 'Edmundo Boulanger'}, { id: null, name: 'Freddie Lanclos'}, { id: null, name: 'Gregory Lickteig'}, { id: null, name: 'Gwendolyn Cuadra'}, { id: null, name: 'Krystal Brosnahan'}, { id: null, name: 'Lahoma Pagani'}, { id: null, name: 'Senaida Risk'}, { id: null, name: 'Valarie Lopes'}];
friends.sort(function(a, b) {
// check id values are equal even the null
if (a.id === b.id) {
// compare the name property in that case
if (a.name < b.name) return -1;
if (a.name > b.name) return 1;
return 0;
}
// else check for null
if (a.id === null) return 1;
if (b.id === null) return -1;
// if both are different return the difference to sort based on that
return a.id - b.id
});
console.log(friends);
答案 2 :(得分：0)
您可以使用算术计算来确定位置：
1或-1并进行比较。
var data=[{id:1122,name:"Adrianna Budzinski"},{id:3785,name:"Amy Divine"},{id:1555,name:"Gale Purdue"},{id:1920,name:"Rex Feng"},{id:2010,name:"Samella Vizcaino"},{id:null,name:"Bethanie Weaver"},{id:null,name:"Celesta Gullo"},{id:null,name:"Darrick Fort"},{id:null,name:"Edmundo Boulanger"},{id:null,name:"Freddie Lanclos"},{id:null,name:"Gregory Lickteig"},{id:null,name:"Gwendolyn Cuadra"},{id:null,name:"Krystal Brosnahan"},{id:null,name:"Lahoma Pagani"},{id:null,name:"Senaida Risk"},{id:null,name:"Valarie Lopes"}];
data.sort(function(a,b){
var _id1 = a.id? 1: -1;
var _id2 = b.id? 1: -1;
var rank = _id1 >_id2 ? -10: _id1< _id2 ? 10 : 0
var v = a.name > b.name? 1: a.name < b.name? -1: 0;
return rank + v;
});
document.getElementById("result").innerHTML = JSON.stringify(data,0,4)<pre id="result"></pre>