问题:如何计算列表中的唯一值
因此,我试图制作一个程序来询问用户输入并将值存储在数组/列表中。
然后,当输入空白行时,它将告诉用户这些值中有多少是唯一的。
我出于现实原因而不是问题集来构建它。
enter: happy
enter: rofl
enter: happy
enter: mpg8
enter: Cpp
enter: Cpp
enter:
There are 4 unique words!
我的代码如下:
# ask for input
ipta = raw_input("Word: ")
# create list
uniquewords = []
counter = 0
uniquewords.append(ipta)
a = 0 # loop thingy
# while loop to ask for input and append in list
while ipta:
ipta = raw_input("Word: ")
new_words.append(input1)
counter = counter + 1
for p in uniquewords:
..这就是到目前为止我所获得的一切。
我不确定如何计算列表中单词的唯一数量?
如果有人可以发布解决方案,以便我可以学习它,或者至少告诉我它会是多么棒,谢谢!
So I’m trying to make this program that will ask the user for input and store the values in an array / list.
Then when a blank line is entered it will tell the user how many of those values are unique.
I’m building this for real life reasons and not as a problem set.
enter: happy
enter: rofl
enter: happy
enter: mpg8
enter: Cpp
enter: Cpp
enter:
There are 4 unique words!
My code is as follows:
# ask for input
ipta = raw_input("Word: ")
# create list
uniquewords = []
counter = 0
uniquewords.append(ipta)
a = 0 # loop thingy
# while loop to ask for input and append in list
while ipta:
ipta = raw_input("Word: ")
new_words.append(input1)
counter = counter + 1
for p in uniquewords:
..and that’s about all I’ve gotten so far.
I’m not sure how to count the unique number of words in a list?
If someone can post the solution so I can learn from it, or at least show me how it would be great, thanks!
回答 0
另外,使用collections.Counter重构代码:
from collections import Counter
words = ['a', 'b', 'c', 'a']
Counter(words).keys() # equals to list(set(words))
Counter(words).values() # counts the elements' frequency
输出:
['a', 'c', 'b']
[2, 1, 1]
In addition, use collections.Counter to refactor your code:
from collections import Counter
words = ['a', 'b', 'c', 'a']
Counter(words).keys() # equals to list(set(words))
Counter(words).values() # counts the elements' frequency
Output:
['a', 'c', 'b']
[2, 1, 1]
回答 1
您可以使用集合删除重复项,然后使用len函数计算集合中的元素:
len(set(new_words))
You can use a set to remove duplicates, and then the len function to count the elements in the set:
len(set(new_words))
回答 2
values, counts = np.unique(words, return_counts=True)
values, counts = np.unique(words, return_counts=True)
回答 3
使用一组:
words = ['a', 'b', 'c', 'a']
unique_words = set(words) # == set(['a', 'b', 'c'])
unique_word_count = len(unique_words) # == 3
有了这个,您的解决方案就可以很简单:
words = []
ipta = raw_input("Word: ")
while ipta:
words.append(ipta)
ipta = raw_input("Word: ")
unique_word_count = len(set(words))
print "There are %d unique words!" % unique_word_count
Use a set:
words = ['a', 'b', 'c', 'a']
unique_words = set(words) # == set(['a', 'b', 'c'])
unique_word_count = len(unique_words) # == 3
Armed with this, your solution could be as simple as:
words = []
ipta = raw_input("Word: ")
while ipta:
words.append(ipta)
ipta = raw_input("Word: ")
unique_word_count = len(set(words))
print "There are %d unique words!" % unique_word_count
回答 4
aa="XXYYYSBAA"
bb=dict(zip(list(aa),[list(aa).count(i) for i in list(aa)]))
print(bb)
# output:
# {'X': 2, 'Y': 3, 'S': 1, 'B': 1, 'A': 2}
aa="XXYYYSBAA"
bb=dict(zip(list(aa),[list(aa).count(i) for i in list(aa)]))
print(bb)
# output:
# {'X': 2, 'Y': 3, 'S': 1, 'B': 1, 'A': 2}
回答 5
对于ndarray,有一个称为unique的numpy方法:
np.unique(array_name)
例子:
>>> np.unique([1, 1, 2, 2, 3, 3])
array([1, 2, 3])
>>> a = np.array([[1, 1], [2, 3]])
>>> np.unique(a)
array([1, 2, 3])
对于系列,有一个函数调用value_counts():
Series_name.value_counts()
For ndarray there is a numpy method called unique:
np.unique(array_name)
Examples:
>>> np.unique([1, 1, 2, 2, 3, 3])
array([1, 2, 3])
>>> a = np.array([[1, 1], [2, 3]])
>>> np.unique(a)
array([1, 2, 3])
For a Series there is a function call value_counts():
Series_name.value_counts()
回答 6
ipta = raw_input("Word: ") ## asks for input
words = [] ## creates list
unique_words = set(words)
ipta = raw_input("Word: ") ## asks for input
words = [] ## creates list
unique_words = set(words)
回答 7
尽管集合是最简单的方法,但是您也可以使用some_dict.has(key)
字典并仅使用唯一的键和值来填充字典。
假设您已经填充words[]
了用户的输入,请创建一个字典,将列表中的唯一单词映射到数字:
word_map = {}
i = 1
for j in range(len(words)):
if not word_map.has_key(words[j]):
word_map[words[j]] = i
i += 1
num_unique_words = len(new_map) # or num_unique_words = i, however you prefer
Although a set is the easiest way, you could also use a dict and use some_dict.has(key)
to populate a dictionary with only unique keys and values.
Assuming you have already populated words[]
with input from the user, create a dict mapping the unique words in the list to a number:
word_map = {}
i = 1
for j in range(len(words)):
if not word_map.has_key(words[j]):
word_map[words[j]] = i
i += 1
num_unique_words = len(new_map) # or num_unique_words = i, however you prefer
回答 8
使用熊猫的其他方法
import pandas as pd
LIST = ["a","a","c","a","a","v","d"]
counts,values = pd.Series(LIST).value_counts().values, pd.Series(LIST).value_counts().index
df_results = pd.DataFrame(list(zip(values,counts)),columns=["value","count"])
然后,您可以以任何所需的格式导出结果
Other method by using pandas
import pandas as pd
LIST = ["a","a","c","a","a","v","d"]
counts,values = pd.Series(LIST).value_counts().values, pd.Series(LIST).value_counts().index
df_results = pd.DataFrame(list(zip(values,counts)),columns=["value","count"])
You can then export results in any format you want
回答 9
怎么样:
import pandas as pd
#List with all words
words=[]
#Code for adding words
words.append('test')
#When Input equals blank:
pd.Series(words).nunique()
它返回列表中有多少个唯一值
How about:
import pandas as pd
#List with all words
words=[]
#Code for adding words
words.append('test')
#When Input equals blank:
pd.Series(words).nunique()
It returns how many unique values are in a list
回答 10
以下应该工作。lambda函数过滤掉重复的单词。
inputs=[]
input = raw_input("Word: ").strip()
while input:
inputs.append(input)
input = raw_input("Word: ").strip()
uniques=reduce(lambda x,y: ((y in x) and x) or x+[y], inputs, [])
print 'There are', len(uniques), 'unique words'
The following should work. The lambda function filter out the duplicated words.
inputs=[]
input = raw_input("Word: ").strip()
while input:
inputs.append(input)
input = raw_input("Word: ").strip()
uniques=reduce(lambda x,y: ((y in x) and x) or x+[y], inputs, [])
print 'There are', len(uniques), 'unique words'
回答 11
我会自己使用一套,但这是另一种方式:
uniquewords = []
while True:
ipta = raw_input("Word: ")
if ipta == "":
break
if not ipta in uniquewords:
uniquewords.append(ipta)
print "There are", len(uniquewords), "unique words!"
I’d use a set myself, but here’s yet another way:
uniquewords = []
while True:
ipta = raw_input("Word: ")
if ipta == "":
break
if not ipta in uniquewords:
uniquewords.append(ipta)
print "There are", len(uniquewords), "unique words!"
回答 12
ipta = raw_input("Word: ") ## asks for input
words = [] ## creates list
while ipta: ## while loop to ask for input and append in list
words.append(ipta)
ipta = raw_input("Word: ")
words.append(ipta)
#Create a set, sets do not have repeats
unique_words = set(words)
print "There are " + str(len(unique_words)) + " unique words!"
ipta = raw_input("Word: ") ## asks for input
words = [] ## creates list
while ipta: ## while loop to ask for input and append in list
words.append(ipta)
ipta = raw_input("Word: ")
words.append(ipta)
#Create a set, sets do not have repeats
unique_words = set(words)
print "There are " + str(len(unique_words)) + " unique words!"
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