使用字典在Python中出现的次数

Question

我有以下练习:

文本在单行中给出。对于文本中的每个单词，计算其在其前面出现的次数。为简单起见，我们假设输入中没有数字、标点符号或特殊符号。连续的单词之间用一个或多个空格分隔。

练习明确要求使用字典。

输入:

the block on the block on the block on the floor

输出:

0 0 0 1 1 1 2 2 2 3 0

目前的代码:

my_words={} 
s = input()
nums_str = s.split()
count=0
for x in nums_str:
my_words.update({count:0})
count= count+1

the block on the block on the block on the floor
{0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0, 6: 0, 7: 0, 8: 0, 9: 0, 10: 0}

输出一个字典，每个单词的索引是一个数字和一个零。

这个想法是用它来记录每个单词在它自己之前的位置。

这就是我不懂的地方如何记录每个单词在自己前面的位置

Answer 1

单靠字典是不够的。如果你添加一个列表，你可以在那里收集当前字数。

目的是检查单词是否在字典中。如果没有，则将其添加并设置为0。如果它在那里，那么增加1的值。然后使用当前添加或更新的值。

s = 'the block on the block on the block on the floor'
words = {}
counts = []
for word in s.split():
if word not in words:
words[word] = 0
else:
words[word] += 1
counts.append(words[word])
print(counts)

Result:[0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 0]

---

如果您不需要为进一步的任务收集输出，您可以删除列表并仅打印当前值。

s = 'the block on the block on the block on the floor'
words = {}
for word in s.split():
if word not in words:
words[word] = 0
else:
words[word] += 1
print(words[word], end=' ')

---

相同的方法与字典的get方法和默认值。(你可以在mousetail的回答中看到类似的方法)

words = {}
for word in s.split():
words[word] = words.get(word, -1) + 1
print(words[word], end=' ')

Answer 2

为了实现这一点，您需要在每次迭代后获取counter的值。另外，为了使这段代码在任何地方都可以重用，您可以用它创建一个函数，如下所示:

def dynamic_counter(sequence, delimiter=' '):
word_dict, count_list = {}, []
for word in sequence.split(delimiter):
count_list.append(word_dict.get(word, 0))
word_dict[word] = word_dict.get(word, 0) + 1
return count_list

sentence = "the block on the block on the block on the floor"
print(*dynamic_counter(sentence))

Answer 3

用简单的字典解决方案:

words = input().split()
count = {}
for word in words:
print(count.get(word, 0), end=" ") # print earlier words
count[word]=count.get(word, 0) + 1 # increase the counter

Answer 4

不清楚为什么你有用字典。没有字典的帮助，您可以这样做:

mystring = 'the block on the block on the block on the floor'
mywords = mystring.split()
outlist = [mywords[:i].count(w) for i, w in enumerate(mywords)]
print(outlist)

如果你需要一本问题中所示格式的字典，那么:

mystring = 'the block on the block on the block on the floor'
mywords = mystring.split()
mydict = {k: v for k, v in enumerate([mywords[:i].count(w) for i, w in enumerate(mywords)])}
print(mydict)

Answer 5

代码非常简单。这个练习是关于学习字典的，所以建议使用它。

words = input().split()
d = {}
for word in words:
if word in d:
d[word]+=1
else:
d[word] = 0
print(d[word], end = " ")