Python 嵌套用于循环、迭代、计算项目的多次出现次数
这是一段代码,在底部,在代码的“注释掉”部分,有一行让我感到惊讶。请看一看:Python 嵌套用于循环、迭代、计算项目的多次出现次数,python,loops,nested,Python,Loops,Nested,这是一段代码,在底部,在代码的“注释掉”部分,有一行让我感到惊讶。请看一看: #!/path/to/python #-*- coding: utf-8 -*- def frequencer(sliced): podium = [] for item in sliced: scat = len(sliced) print ("le the first scat for the word ' {0} ' is '{1} '.".forma
#!/path/to/python
#-*- coding: utf-8 -*-
def frequencer(sliced):
podium = []
for item in sliced:
scat = len(sliced)
print ("le the first scat for the word ' {0} ' is '{1} '.".format(item, scat))
for indice in range (len(sliced)):
print("indice = ", indice)
print("sliced[indice]",sliced[indice])
if sliced[indice] == item:
print ("sliced[indice] is equal to ' {0} ', identical to ' {1} ' item.".format(sliced[indice], item))
scat -= 1
print("scat is equal to ' {0} '.".format(scat))
print("scat exdented: ", scat)
frequence = len(sliced) - scat
print("frequence: ", frequence)
podium += [frequence]
print("podium: ", podium)
print("podium: ", podium)
return(max(podium))
print(frequencer( ['Here', 'is', 'a', 'line', 'like', 'sparkling', 'wine', 'Line', 'up', 'now', 'behind', 'the', 'cow']))
'''
le the first scat for the word ' line ' is '13 '.
indice = 0
sliced[indice] Here
indice = 1
sliced[indice] is
indice = 2
sliced[indice] a
indice = 3
sliced[indice] line
sliced[indice] is equal to ' line ', identical to ' line ' item.
scat is equal to ' 12 '.
indice = 4
sliced[indice] like
indice = 5
sliced[indice] sparkling
indice = 6
sliced[indice] wine
indice = 7
sliced[indice] Line <-- *WHY IS THIS NOT CONSIDERED EQUAL TO "line"?*
indice = 8
sliced[indice] up
indice = 9
sliced[indice] now
indice = 10
sliced[indice] behind
indice = 11
sliced[indice] the
indice = 12
sliced[indice] cow
scat exdented: 12
frequence: 1
podium: [1, 1, 1, 1]
'''
我的问题是:
项目行在列表中表示了2次,我确信scat=11,frequency=2
我尝试过许多不同的缩进,但主要的兴趣是我没有能力跟随程序命令机器的操作过程
为了说明这一点,我尝试打印了许多步骤,但我确实需要一些进一步的澄清。请帮忙 Python字符串比较区分大小写。所以'A'='A'是假的。如果要进行不区分大小写的比较,应该使用lower或upper方法将字符串设置为小写或大写。所以'A'。lower=='A'是真的 在您的情况下,由于它们是多个字符串,您可能希望在这两个字符串上都使用lower,例如sliced[indice].lower==item.lower。您还可以在开始之前将整个列表转换为小写,从而完全避免问题,如下所示:
slicedlow = [item.lower() for item in sliced]
但是,可以使用集合将整个算法转换为几行。计数器对象:
甚至一行:
from collections import Counter
maxcount = max(Counter(item.lower() for item in sliced).values())
@逻辑。非常感谢您对样式的修改。正如你所猜测的,读英语和写作对我来说并不容易,但我喜欢这样。是的!我是漫不经心的。黑猫:谢谢,如果这是正确的答案,请把它标记成这样,这样其他人就可以找到它。
from collections import Counter
maxcount = max(Counter(item.lower() for item in sliced).values())