根据单词所属的类别集重新排列字符串中的单词

Question

如何根据字符串所属的类别重新排列字符串？假设我有这些套装：

dogs = {'husky', 'chihuahua', 'labrador', 'beagle'} 
flowers = {'dandelion', 'rose', 'tulip'} 
colours = {'blue', 'yellow', 'green', 'red', 'pink'}

然后假设我想输入一个字符串，并根据它们的类别重新排列单词。

'husky tulip red orange'

将成为

'red orange husky tulip'

顺序是先是颜色，然后是狗，然后是花。也许可以按顺序创建一个类别列表？不太确定我会怎么做这个

Answer 1

使用带有sorted:的键函数

def ref(s):
dogs = {'husky', 'chihuahua', 'labrador', 'beagle'} 
flowers = {'dandelion', 'rose', 'tulip'} 
colours = {'blue', 'yellow', 'green', 'red', 'pink', 'orange'}
if s in colours: rtr=-3
elif s in dogs: rtr=-2
elif s in flowers: rtr=-1
else: rtr=0   # this puts words not found at end of string
return rtr 
s='husky tulip red orange'
>>> ' '.join(sorted(s.split(), key=ref))
red orange husky tulip

更多Python(更容易扩展(是这样做的：

def ref(s):
dogs = {'husky', 'chihuahua', 'labrador', 'beagle'} 
flowers = {'dandelion', 'rose', 'tulip'} 
colours = {'blue', 'yellow', 'green', 'red', 'pink', 'orange'}
key_t=(colours, dogs, flowers)
try: 
return next(i for i, v in enumerate(key_t) if s in v)
except StopIteration:
return -1. # this puts words not found at beginning of string
# or use the default argument version of next:
# return next((i for i, v in enumerate(key_t) if s in v), -1)

并以同样的方式使用该键函数。

您还可以通过使用chain将集合链接到一个可迭代的集合中来迭代集合：

>>> from itertools import chain
>>> [e for e in chain(colours,dogs,flowers) if e in s.split()]
['orange', 'red', 'husky', 'tulip']

哪个更快或更好取决于字符串的大小和集合的大小。此外，如果您想进行二次排序(例如在各个类别中进行词典编纂(，则需要使用sorted方法。

Answer 2

试试这个

#Define all the categories
dogs = ['husky', 'chihuahua', 'labrador', 'beagle']
flowers = ['dandelion', 'rose', 'tulip']
colours = ['blue', 'yellow', 'green', 'red', 'pink', 'orange']
#The Input String
outOfOrder = "husky tulip red orange"
#Split up the string into an array which each word seperated
outOfOrderArray = outOfOrder.split()
#Array to hold all words of each category
orderedArray = [[], [], [], []]
#loop through all the words in the array
for word in outOfOrderArray:
#Check if the word is in each category.
if word in dogs:
orderedArray[2].append(word)
elif word in flowers:
orderedArray[1].append(word)
elif word in colours:
orderedArray[0].append(word)
#If its not in the array, do whatever you want with it. I jsut stuck them at the end.
else:
orderedArray[3].append(word)
orderedString = ""
#Combine all the words in ordered Array to create a final string
for category in orderedArray:
for word in category:
orderedString = orderedString + word + " "
print(orderedString)

Answer 3

您可以用1键将鲜花推到后面，用-1键将颜色拉到前面(其他所有东西都会用0键放在中间(：

>>> ' '.join(sorted(s.split(), key=lambda w: (w in flowers) - (w in colours)))
'red husky orange tulip'

注意，"；橙色"；不在你的任何类别中，这就是为什么它最终与"；哈士奇"；。

一种更通用的方法是将花朵向后推得最远，将狗向后推得最少，将颜色向后推得最小：

>>> ' '.join(sorted(s.split(), key=lambda w: (w in flowers, w in dogs, w in colours)))
'orange red husky tulip'

Answer 4

只需按以下顺序从集合中提取匹配字符串：

>>>[i for i in list(colours)+list(flowers)+list(dogs) if i in input_list]
['red', 'orange', 'tulip', 'husky']

无需执行排序或其他操作。

如果您最初将它们定义为列表而不是集合，则会更简单。此外，它将保留您将放入列表的顺序，而不是对所有项目进行集体排序：

dogs = ['husky', 'chihuahua', 'labrador', 'beagle']
flowers = ['dandelion', 'rose', 'tulip'] 
colours = ['blue', 'yellow', 'green', 'red', 'pink', 'orange']
input_list = 'husky tulip red orange pink blue'.split()
[i for i in colours+flowers+dogs if i in input_list]

输出：

['blue', 'red', 'pink', 'orange', 'tulip', 'husky'] # note the ordering

p.S.对我来说似乎是最具蟒蛇性、最具时空效率、最具可扩展性和最快的方法。