6

社区。我需要接受多个以逗号分隔的输入来生成信息摘要（具体来说，每个小组/项目有多少不同的员工参与）？该程序以字符串的形式接收员工、经理和组。

我正在使用 anytree python 库来搜索/计算每组每个员工的出现次数。但是，该程序一次只接受一个值/单元格，而不是多个值。

这是树结构以及我如何接受输入值？

Press q to exit, Enter your data: Joe
Press q to exit, Enter your data: Manager1
Press q to exit, Enter your data: Group1
Press q to exit, Enter your data: Charles 
Press q to exit, Enter your data: Manager1
Press q to exit, Enter your data: Group2
Press q to exit, Enter your data: Joe
Press q to exit, Enter your data: Manager3
Press q to exit, Enter your data: Group1
Press q to exit, Enter your data: Charles
Press q to exit, Enter your data: Manager3
Press q to exit, Enter your data: Group1
Press q to exit, Enter your data: Joe
Press q to exit, Enter your data: Manager5
Press q to exit, Enter your data: Group2
Press q to exit, Enter your data: q
Employee   No of groups
   JOE       2
   CHARLES       2
Group
├── GROUP1
│   ├── JOE
│   │   └── MANAGER1
│   ├── JOE
│   │   └── MANAGER3
│   └── CHARLES
│       └── MANAGER3
└── GROUP2
    ├── CHARLES
    │   └── MANAGER1
    └── JOE
        └── MANAGER5

我需要有关此代码的帮助，以便它可以接受逗号分隔的值；例如，一次输入Joe、Manager1、Group1。

import anytree

from anytree import Node, RenderTree, LevelOrderIter, LevelOrderGroupIter, PreOrderIter

import sys

# user input
io=''
lst_input = []
while (io!='q'):
    io=input('Press q to exit, Enter your data: ')
    if io!='q':
        lst_input.append(io.upper())

# change list in to matrix
lst=[]
for i in range(0, len(lst_input), 3):
    lst.append(lst_input[i:i + 3])

lst

# create tree structure from lst
group = Node('Group')
storeGroup = {}
for i in range(len(lst)):
    if lst[i][2] in [x.name for x in group.children]: # parent already exist, append childrens
        storeGroup[lst[i][0]] = Node(lst[i][0], parent=storeGroup[lst[i][2]])
        storeGroup[lst[i][1]] = Node(lst[i][1], parent=storeGroup[lst[i][0]])
    else: # create parent and append childreds
        storeGroup[lst[i][2]] = Node(lst[i][2], parent=group)
        storeGroup[lst[i][0]] = Node(lst[i][0], parent=storeGroup[lst[i][2]])
        storeGroup[lst[i][1]] = Node(lst[i][1], parent=storeGroup[lst[i][0]])


store = {}
for children in LevelOrderIter(group, maxlevel=3):
    if children.parent!=None and children.parent.name!='Group':
        if children.name not in store:
            store[children.name] = {children.parent.name}
        else:
            store[children.name] = store[children.name] | {children.parent.name}

print('Employee', '  No of groups')
for i in store:
    print('   '+i+'      ', len(store[i]))


for pre,fill, node in RenderTree(group):
    print('{}{}'.format(pre,node.name))


谢谢！欢迎任何想法。

4

3 回答 3

5

利用解包来提取元素。那么 if 语句可以这样重写。

if io!='q':
    name, role, grp = io.upper(). split(',')
    lst_input.append([name,role, grp])

您还需要lst.append(lst_input[i:i + 3])在 for 循环中更改为此。

lst.append(lst_input[0][i:i + 3])
于 2022-02-14T10:02:58.027 回答
3

我相信一种方法是：

name, role, grp = io.upper().split(',')

对于输入，例如Joe, Manager1, Group1

Python 3.7.9 (v3.7.9:13c94747c7, Aug 15 2020, 01:31:08)
[Clang 6.0 (clang-600.0.57)] on darwin
Type "help", "copyright", "credits" or "license" for more information.

>>> name, role, grp = input("Press q to exit, Enter your data:").split(",")
Press q to exit, Enter your data:Joe, Manager1, Group1
>>> name
'Joe'
>>> role
' Manager1'
>>> grp
' Group1'

这对你有用吗？

于 2022-02-13T12:47:12.363 回答
2

您的lst输出类似于：

[['Joe', 'Manager1', 'Group1'], ['Charles', 'Manager1', 'Group2'], ['Joe', 'Manager3', 'Group1'], ['Charles', 'Manager3', 'Group1'], ['Joe', 'Manager5', 'Group2']]

下面的代码lst以更好的性能产生相同的输出：

import anytree

from anytree import Node, RenderTree, LevelOrderIter, LevelOrderGroupIter, PreOrderIter

import sys

# user input
lst = []
while True:
   io=input('Press q to exit, Enter your data: ')
   if io!='q':
      lst.append(io.upper().split(',')) 
   else:
      break
print(lst)

# create tree structure from lst
group = Node('Group')
storeGroup = {}
for i in range(len(lst)):
    if lst[i][2] in [x.name for x in group.children]: # parent already exist, append childrens
        storeGroup[lst[i][0]] = Node(lst[i][0], parent=storeGroup[lst[i][2]])
        storeGroup[lst[i][1]] = Node(lst[i][1], parent=storeGroup[lst[i][0]])
    else: # create parent and append childreds
        storeGroup[lst[i][2]] = Node(lst[i][2], parent=group)
        storeGroup[lst[i][0]] = Node(lst[i][0], parent=storeGroup[lst[i][2]])
        storeGroup[lst[i][1]] = Node(lst[i][1], parent=storeGroup[lst[i][0]])


store = {}
for children in LevelOrderIter(group, maxlevel=3):
    if children.parent!=None and children.parent.name!='Group':
        if children.name not in store:
            store[children.name] = {children.parent.name}
        else:
            store[children.name] = store[children.name] | {children.parent.name}

print('Employee', '  No of groups')
for i in store:
    print('   '+i+'      ', len(store[i]))


for pre,fill, node in RenderTree(group):
    print('{}{}'.format(pre,node.name))

在您的原始代码中定义许多不必要的列表会破坏您的性能。甚至避免进行额外的比较或定义不必要的变量。请注意，我们do ... while在Python. 因此，我们使用break避免重复比较。

于 2022-02-20T08:43:58.173 回答