我试图通过获取一个百分比来分割列表,然后从主列表中随机地抓取元素到另外两个列表中.trainingSet是左上方列表.当我生成一个随机索引来挑选时,我遇到了一个问题.这段代码适用于一个小列表,但当我使用(len(rawRatings)= 1000)时,它不起作用.
错误:
File "/Applications/WingIDE.app/Contents/MacOS/src/debug/tserver/_sandbox.py", line 1, in <module>
# Used internally for debug sandbox under external interpreter
File "/Applications/WingIDE.app/Contents/MacOS/src/debug/tserver/_sandbox.py", line 29, in partitionRankings
File "/Users/rderickson9/anaconda/lib/python2.7/random.py", line 241, in randint
return self.randrange(a, b+1)
File "/Users/rderickson9/anaconda/lib/python2.7/random.py", line 217, in randrange
raise ValueError, "empty range for randrange() (%d,%d, %d)" % (istart, istop, width)
ValueError: empty range for randrange() (0,0, 0)
raw Ratings是一个列表,testPercent是一个浮点数.
恩.
rawRatings = [(123,432,4),(23,342,3),(23,123,5),(234,523,3),(34,23,1), (12,32,4)]
testPercent = .2
partitionRankings(rawRatings, testPercent)
[(23,123,5),(234,523,3),(34,23,1),(123,432,4),(12,32,4)],[(23,342,3)]
def partitionRankings(rawRatings, testPercent):
testSet = []
trainingSet = [] …我正在尝试使用嵌套列表中的字典来计算单词的频率.每个嵌套列表都是分成每个单词的句子.另外,我想删除句子开头的专有名词和小写单词.是否有可能获得专有名词？
x = [["Hey", "Kyle","are", "you", "doing"],["I", "am", "doing", "fine"]["Kyle", "what", "time" "is", "it"]
from collections import Counter
def computeFrequencies(x):
count = Counter()
for listofWords in L:
for word in L:
count[word] += 1
return count
它返回一个错误:unhashable type:'list'
我想在没有字典周围的Counter()的情况下返回这个:
{"hey": 1, "how": 1, "are": 1, "you": 1, "doing": 2, "i": , "am": 1, "fine": 1, "what": 1, "time": 1, "is": 1, "it": 1}
我正在尝试组合多个元组,但不知道如何.我知道如何创建一个元组,但我不太确定如何将它们组合在一起.我想继续追加(实际上不是追加,因为我不想要一个列表).我从每一行中拉出一根弦并将所有元组放在一起
x = (132, 534, 4)
y = (345, 531, 1)
z = (212, 421, 5)
我想要的是什么
(132, 534, 4), (345, 531, 1), (212, 421, 5)