使用Python从列表中进行多项选择

Question

我创建了一个很大的服务器名称列表。我想将此作为菜单呈现给用户并创建一个新列表，该列表仅包含用户选择的名称/多个选项。

BigList = ['server1','server2','server3','server4']
while (i < len(BigList)):
    i =+ 1
print "%d. %s Return to main menu" % (i+1,BigList)
menu = raw_input("Enter the selection with comma:")
menu = menu.split(",")
return menu

当用户从菜单中选择多个选项时，我得到的列表是一个数字列表，而不是要返回的实际服务器名称。此外，没有错误处理，例如数字是否在菜单中有效。

感谢任何帮助，我是Python的新手，试图了解更多信息。

Answer 1

看到你还没有得到答案，我只想举例说明你认为你可能想要实现的目标：

BigList = ['server1','server2','server3','server4']

# here we use enumerate to generate the index i for the current item
# we pass in a temp list with an extra entry concatenated so that we
# dont need to duplicate the print expression
input_list = BigList + ['Return to main menu']
for i, item in enumerate(input_list):
    print "%d. %s" % (i, item)

# get user input
menu_input = raw_input("Enter the selection with comma:")

# the list of selection indexes 
menu_selection_indexes = []

# sanitize and store the indexes from user input string
for i in menu_input.split(','):
    # could print/throw errors on bad input
    # here we choose to discard bad list items silently
    try:
            # convert the string e.g. "2" into an integer type
            # so that we can use it to index into our menu list
        val = int(i.strip())
    except ValueError, ve:
            # some strings (e.g. "a") cannot be converted to an integer
            # a ValueError exception is thrown if this is attempted
            # we catch this exception and just ignore this value,
            # skipping this value continuing on with the next for-loop item
        continue
    # ignore indexes that exceeed the acceptable input list size
    if val > len(input_list):
        continue
    menu_selection_indexes.append(val)

# print indexes
print menu_selection_indexes

# if the last possible input index was chosen, we return to menu
if len(BigList) in menu_selection_indexes:
    if not len(menu_selection_indexes) == 1:
        print "warning: mixed commands"
    print "will return to main menu"

else:
    # list of items selected, using list comprehensions
    menu_selection_names = [input_list[i] for i in menu_selection_indexes]

    # print names
    print menu_selection_names

Answer 2

我想你想打印每个元素旁边有一个索引。 enumerate function返回一个可迭代对象。这个可迭代对象在每次迭代时返回一个元组，由一个索引（从0开始）和列表中该索引处的元素组成。
然后，在第二次，您可以遍历用户选择的每个索引，并将revelant元素添加到selected列表。

BigList = ['server1','server2','server3','server4']
# Show every index
for index, element in enumerate(BigList):
    print "%d. %s" % (index, element)

print "%d. %s Return to main menu" % (i+1,BigList)

menu = raw_input("Enter the selection with comma:")
menu.split(",")
selected = []
for data in menu:
     try:
          index = int(data.strip())    
          selected.append(BigList[index])
     except ValueError:
          pass  # Need to manage this case, where it's not an integer

return selected

Answer 3

返回/打印包含QListWidget中所选项目的列表：

import sys

from PyQt5.QtWidgets import (QApplication,  QGridLayout, QLabel, QWidget, QListWidget, QAbstractItemView, QMainWindow)

class Main(QMainWindow):

    def __init__(self, parent = None):
        QMainWindow.__init__(self,parent)

        self.initUI()

    def initUI(self):

        self.setWindowTitle('Input dialog')
        win=QWidget()
        self.setCentralWidget(win)
        mainLayout = QGridLayout()

        self.std = QListWidget()
        self.std.setSelectionMode(QAbstractItemView.MultiSelection)
        self.std.addItems(["AB", "EF", "JK"])
        self.std.itemClicked.connect(self.selectionchange)

        self.stdl = QLabel(self)
        self.stdl.setText('select')  

        mainLayout.addWidget(self.stdl,  0, 0)
        mainLayout.addWidget(self.std, 0, 1)

        win.setLayout(mainLayout)
        win.show()

    def selectionchange(self):
        se=[]
        for i in self.std.selectedItems():
            m=i.text()
            se.append(m)
        print(se)

def main():

    app = QApplication(sys.argv)

    main = Main()
    main.show()

    sys.exit(app.exec_())
if __name__ == "__main__":
    main()