因此,我试图使这个程序,将要求用户输入,并将值存储在一个数组/列表。
然后,当输入空行时,它会告诉用户这些值中有多少是唯一的。
我做这个是出于现实生活的原因,而不是作为习题集。
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:
..到目前为止我就知道这么多
我不知道如何计算一个列表中唯一的单词数?
如果有人可以发布解决方案,这样我就可以从中学习,或者至少向我展示它是如何伟大的,谢谢!
另一种方法是用熊猫
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,您可以将结果导出为所需的任何格式
另外,使用集合。计数器重构你的代码:
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]
使用集合:
words = ['a', 'b', 'c', 'a']
unique_words = set(words) # == set(['a', 'b', 'c'])
unique_word_count = len(unique_words) # == 3
有了这个,你的解决方案可以很简单:
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
另一种方法是用熊猫
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,您可以将结果导出为所需的任何格式