blocks|key|1858355|text|如果您正在做家庭作业，那么这不会有什么帮助--但是如果您正在寻找一种快速实现目标的方法，那么来自collections模块的collections函数应该是有意义的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1858356|text+=+'You+can+have+data+without+information,+but+you+cannot+have+information+without+data.'
from+collections+import+Counter
Counter(text.lower())

>>+Counter({+"y"+:+2,+
+++++++++++++"o"+:+9,+
+++++++++++++"u"+:+5,
+++++++++++++...etc|code-block|syntax|javascript|1858357|entityMap^0|1C|B|1Q|B|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

If you're doing a homework assignment, then this isn't going to help - but if you're looking for a quick way to achieve your goal, then the <code>Counter</code> function from the <code>collections</code> module should be of interest:
<pre><code>text = 'You can have data without information, but you cannot have information without data.'
from collections import Counter
Counter(text.lower())

&gt;&gt; Counter({ &quot;y&quot; : 2, 
 &quot;o&quot; : 9, 
 &quot;u&quot; : 5,
 ...etc
</code></pre>

blocks|key|518129|text|我想你要求出现在字符串中，第一个例子是列表中单词的出现，另一个例子是使用字典在字符串中出现字母，但你也可以在那里使用计数器！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|518130|第一个示例：|offset|length|style|BOLD|518131|+++from+collections+import+Counter

+++text+=+'You+can+have+data+without+information,+but+you+cannot+have+information+without+data.'
+++text+=+text.split()
+++text+=+Counter(text)
+++print(text)|code-block|syntax|javascript|518132|第二个示例：|518133|++++text+=+'You+can+have+data+without+information,+but+you+cannot+have+information+
without+data.'
++++text+=+text.lower()
++++text_dictionary+=+{}
++++for+element+in+text:
++++++++if+element+not+in+text_dictionary+and+element.isalpha():
++++++++++++text_dictionary[element]+=+1
++++++++else:
++++++++++++if+element.isalpha():
++++++++++++++++text_dictionary[element]+%2B=+1
++++print(text_dictionary)|518134|entityMap^0|0|0|6|0|0|0|6|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

I think you asked for occurences in a string ,first example is occurences of the
words in a list,and the other example is occurences of letters in a string using dictionary but you can also use Counter there!
First Example:
<pre><code> from collections import Counter

 text = 'You can have data without information, but you cannot have information without data.'
 text = text.split()
 text = Counter(text)
 print(text)
</code></pre>
Second Example:
<pre><code> text = 'You can have data without information, but you cannot have information 
without data.'
 text = text.lower()
 text_dictionary = {}
 for element in text:
 if element not in text_dictionary and element.isalpha():
 text_dictionary[element] = 1
 else:
 if element.isalpha():
 text_dictionary[element] += 1
 print(text_dictionary)
</code></pre>

blocks|key|1862533|text|list(set(lowerCaseText)将返回字符串中的所有唯一字符。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1862534|然后，您可以遍历这些唯一字符，并使用lowerCaseText.count(char)获取它们在初始字符串中出现的次数。|1862535|把所有的东西放在一起，你就会得到这样的东西：|1862536|text+=+'You+can+have+data+without+information,+but+you+cannot+have+information+without+data.'
lowerCaseText+=+text.lower()
uniq+=+sort(list(set(lowerCaseText)))
for+char+in+uniq:
++++print(char,+':',+lowerCaseText.count(char))|code-block|syntax|javascript|1862537|entityMap^0|0|N|0|I|P|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|W|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|X|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

<code>list(set(lowerCaseText)</code> will return all of the unique characters in your string.
You can then go through those unique characters and get a count of their occurrences in your initial string by using <code>lowerCaseText.count(char)</code>
Put it all together and you get something like this:
<pre><code>text = 'You can have data without information, but you cannot have information without data.'
lowerCaseText = text.lower()
uniq = sort(list(set(lowerCaseText)))
for char in uniq:
 print(char, ':', lowerCaseText.count(char))
</code></pre>

blocks|key|1858378|text|代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1858379|import+re++++++#This+package+to+remove+space+and+special+char+from+text
import+string++#This+package+to+alphabets


text+=+'You+can+have+data+without+information,+but+you+cannot+have+information+without+data.'
Alpha+=+dict.fromkeys(string.ascii_lowercase,+0)+++#Output+{a:0,b:0+..}

for+t+in+re.sub('[%5EA-Za-z0-9]%2B',+'',+text):++++++++#using+re+avoiding+the+spaces+and+special+characters
++++Alpha[t.lower()]+=+Alpha[t.lower()]%2B1++++++++++#each+time+add+1+on+specific+alphabet
++++
Alpha|code-block|syntax|javascript|1858380|输出：|1858381|{'a':+10,
+'b':+1,
+'c':+2,
+'d':+2,
+'e':+2,
+'f':+2,
+'g':+0,
+'h':+4,
+'i':+6,
+'j':+0,
+'k':+0,
+'l':+0,
+'m':+2,
+'n':+7,
+'o':+9,
+'p':+0,
+'q':+0,
+'r':+2,
+'s':+0,
+'t':+10,
+'u':+5,
+'v':+2,
+'w':+2,
+'x':+0,
+'y':+2,
+'z':+0}|1858382|要获得您的期望输出，{k:v+for+k,v+in+Alpha.items()+if+v>0}输出：|offset|length|style|CODE|1858383|{'a':+10,+'b':+1,+'c':+2,+'d':+2,+'e':+2,+'f':+2,+'h':+4,+'i':+6,+'m':+2,+'n':+7,+'o':+9,+'r':+2,+'t':+10,+'u':+5,+'v':+2,+'w':+2,+'y':+2}|1858384|entityMap^0|0|0|0|0|A|11|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Code:
<pre><code>import re #This package to remove space and special char from text
import string #This package to alphabets


text = 'You can have data without information, but you cannot have information without data.'
Alpha = dict.fromkeys(string.ascii_lowercase, 0) #Output {a:0,b:0 ..}

for t in re.sub('[^A-Za-z0-9]+', '', text): #using re avoiding the spaces and special characters
 Alpha[t.lower()] = Alpha[t.lower()]+1 #each time add 1 on specific alphabet
 
Alpha
</code></pre>
Output:
<pre><code>{'a': 10,
 'b': 1,
 'c': 2,
 'd': 2,
 'e': 2,
 'f': 2,
 'g': 0,
 'h': 4,
 'i': 6,
 'j': 0,
 'k': 0,
 'l': 0,
 'm': 2,
 'n': 7,
 'o': 9,
 'p': 0,
 'q': 0,
 'r': 2,
 's': 0,
 't': 10,
 'u': 5,
 'v': 2,
 'w': 2,
 'x': 0,
 'y': 2,
 'z': 0}
</code></pre>
To get your desire output <code>{k:v for k,v in Alpha.items() if v&gt;0}</code>
output:
<pre><code>{'a': 10, 'b': 1, 'c': 2, 'd': 2, 'e': 2, 'f': 2, 'h': 4, 'i': 6, 'm': 2, 'n': 7, 'o': 9, 'r': 2, 't': 10, 'u': 5, 'v': 2, 'w': 2, 'y': 2}
</code></pre>

blocks|key|518147|text|对于一种不同的做事方式，这里有一个不同而粗略的想法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|518148|所以你可以做的是有一个26个值的数组，每个字母1。然后是第二个数组，有26个值，都从0开始。因此，您可以循环遍历第一个数组，计算在文本中找到该值的次数，然后在第二个数组中设置该值。|518149|A=‘a’，‘b’，to‘z’b=+0,0,0，…0|518150|对于字母表中的每一个字母:b字符值=+text.count(acharacter值)|518151|然后，您可以执行一个for循环，以便以您想要的方式打印东西。|518152|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|O|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

Here’s a different and rough idea for a different way to do things:
So what you could do is have an array of 26 values, 1 for each letter of the alphabet. And then a 2nd array with 26 values, all starting with 0.
So then you can loop through the 1st array, figure out how many times that value is found in the text, and then set that value in the 2nd array.
a = [‘a’, ‘b’, to ‘z’]
b = [0,0,0,…0]
for each letter in the alphabet:
B[character value] = text.count(a[character value])
Then you could do a for loop to get things printed the way you want.

blocks|key|1858405|text|您可以使用collections.Counter。我还使用collections.OrderedDict对值进行排序。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1858406|from+collections+import+Counter,+OrderedDict

text+=+'You+can+have+data+without+information,+but+you+cannot+have+information+without+data.'
lower_case_text+=+text.lower()
lower_case_text+=+''.join(filter(str.isalnum,+lower_case_text))+#+this+line+is+required+for+removing+non+alpha+numeric+characters.
counter+=+Counter(lower_case_text)
counter_ordered=OrderedDict(counter.most_common())
for+k,+v+in+counter_ordered.items():
++++print(k+%2B+"+:+"+%2B+str(v))|code-block|syntax|javascript|1858407|entityMap^0|5|J|T|N|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

You can use <code>collections.Counter</code>. I also used <code>collections.OrderedDict</code> to sort values.
<pre><code>from collections import Counter, OrderedDict

text = 'You can have data without information, but you cannot have information without data.'
lower_case_text = text.lower()
lower_case_text = ''.join(filter(str.isalnum, lower_case_text)) # this line is required for removing non alpha numeric characters.
counter = Counter(lower_case_text)
counter_ordered=OrderedDict(counter.most_common())
for k, v in counter_ordered.items():
 print(k + &quot; : &quot; + str(v))
</code></pre>

blocks|key|518154|text|首先，您需要过滤掉所有不是字符串中的字母的内容。在将字符串转换为dict之前，还需要对字符串进行排序，以便将值按正确的顺序添加。这也是有可能的分类后，但这是更困难的情况下。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|518155|我们使用这一行代码来完成这一任务：|518156|filteredText+=+sorted(''.join(filter(str.isalnum,+text)).lower())|offset|length|style|CODE|518157|(str.isalnum,+text)删除所有非字母，但它作为一个列表出现，所以我们需要再次加入它。|518158|然后，当我们需要检查文本，并检查它是否已经在我们的字典中。如果不是，我们需要用值1来添加它，否则我们就用1来增加。|518159|完整代码：|518160|text+=+'You+can+have+data+without+information,+but+you+cannot+have+information+without+data.'
filteredText+=+sorted(''.join(filter(str.isalnum,+text)).lower())

chars+=+{}

for+char+in+filteredText:
++if+char+in+chars:
++++++chars[char]+%2B=+1
++else+:
++++++chars[char]+=+1

print(chars)|code-block|syntax|javascript|518161|entityMap^0|0|0|0|1T|0|0|J|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Y|8|@$F|Z|G|10|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|11|8|@$F|12|G|13|H|I]]|9|@]|A|$]]|$1|L|3|M|5|6|7|14|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|15|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|16|8|@]|9|@]|A|$S|T]]|$1|U|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|V|$]]

first you need to filter out everything that is not a letter from your string. You also need to preferably sort your string before you convert it into a dict so that the values get added in the right order. It's also possible to sort the dict afterwards but that's more difficult in this case.
We do this using this line of code:
<code>filteredText = sorted(''.join(filter(str.isalnum, text)).lower())</code>
<code>(str.isalnum, text)</code> removes all non letters but it comes out as a list so we need to join it later again.
then when we need to go over the text and check if it is already in our dictionary. If not we need to add it with value 1, otherwise we increment with 1.
Full code:
<pre><code>text = 'You can have data without information, but you cannot have information without data.'
filteredText = sorted(''.join(filter(str.isalnum, text)).lower())

chars = {}

for char in filteredText:
 if char in chars:
 chars[char] += 1
 else :
 chars[char] = 1

print(chars)
</code></pre>

blocks|key|520024|text|很好的问题，这是您的默认解决方案，请告诉我，您在哪里学习python问题！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|520025|from+collections+import+defaultdict
text+=+'You+can+have+data+without+information,+but+you+cannot+have+information+without+data.'

lowerCaseText+=+text.lower()
chars+=+defaultdict(int)

for+char+in+lowerCaseText:
+++if+char.isalpha():++++++#+Only+taking+alphabet+removing+.+,+space
++++++chars[char]+%2B=+1


for+i,+j+in+sorted(chars.items()):
+++print(i,+":",+j)|code-block|syntax|javascript|520026|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Nice problem here is solution for your defaultdict
Please let me know where are you learning python problems!
<pre><code>from collections import defaultdict
text = 'You can have data without information, but you cannot have information without data.'

lowerCaseText = text.lower()
chars = defaultdict(int)

for char in lowerCaseText:
 if char.isalpha(): # Only taking alphabet removing . , space
 chars[char] += 1


for i, j in sorted(chars.items()):
 print(i, &quot;:&quot;, j)
</code></pre>

blocks|key|1858440|text|我们可以利用一个字符也是一个数字:从ascii代码中减去97，给出一个介于0到25之间的数字，我们可以使用它作为列表的索引：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1858441|text+=+'You+can+have+data+without+information,+but+you+cannot+have+information+without+data'.lower()
lst++=+[0]+*+26

#counts+iterations
for+char+in+text:
++++if+not+char.isalpha():+continue
++++lst[ord(char)+-+97]+%2B=+1
+
#prints+the+result:
for+i+in+range(26):
++++if+lst[i]+>+0:
++++++++string+=+chr(i+%2B+97)+%2B+"+:+"+%2B+str(lst[i])
++++++++print(string)

#=>
a+:+10
b+:+1
c+:+2
d+:+2
e+:+2
f+:+2
h+:+4
i+:+6
m+:+2
n+:+7
o+:+9
r+:+2
t+:+10
u+:+5
v+:+2
w+:+2
y+:+2|code-block|syntax|javascript|1858442|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

We can take advantage that a character is also a number: subtracting 97 from the ascii code gives us an number between 0 and 25 that we can use as the index of a list:
<pre><code>text = 'You can have data without information, but you cannot have information without data'.lower()
lst = [0] * 26

#counts iterations
for char in text:
 if not char.isalpha(): continue
 lst[ord(char) - 97] += 1
 
#prints the result:
for i in range(26):
 if lst[i] &gt; 0:
 string = chr(i + 97) + &quot; : &quot; + str(lst[i])
 print(string)

#=&gt;
a : 10
b : 1
c : 2
d : 2
e : 2
f : 2
h : 4
i : 6
m : 2
n : 7
o : 9
r : 2
t : 10
u : 5
v : 2
w : 2
y : 2
</code></pre>

blocks|key|1858444|text|这三个解决方案使用来自string模块的string。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1858445|首先，在迭代时打印：|1858446|from+string+import+ascii_lowercase
text+=+"You+can+have+data+without+information,+but+you+cannot+have+information+without+data."

for+char+in+ascii_lowercase:
++++count+=+0
++++for+letter+in+text.lower():
++++++++if+letter+==+char:
++++++++++++count+%2B=+1
++++if+count:+
++++++++print(f'{char}+:+{count}')|code-block|syntax|javascript|1858447|第二，使用字典使用dict.update()更新计数值。|1858448|from+string+import+ascii_lowercase
text+=+"You+can+have+data+without+information,+but+you+cannot+have+information+without+data."

count+=+dict()
for+char+in+ascii_lowercase:
++++if+char+in+text:
++++++++count.update({char:+text.lower().count(char)})
for+char+in+count:
++++print(f'{char}+:+{count[char]}')|1858449|第三，运用dict理解：|1858450|from+string+import+ascii_lowercase
text+=+"You+can+have+data+without+information,+but+you+cannot+have+information+without+data."
count+=+{char:+text.lower().count(char)+for+char+in+ascii_lowercase+if+char+in+text}
for+char+in+count:+print(f'{char}+:+{count[char]}')|1858451|结果：|1858452|a+:+10
b+:+1
c+:+2
d+:+2
e+:+2
f+:+2
h+:+4
i+:+6
m+:+2
n+:+7
o+:+9
r+:+2
t+:+10
u+:+5
v+:+2
w+:+2
y+:+2|1858453|entityMap^0|B|6|K|6|0|0|0|9|D|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1B|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|1C|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|1D|8|@]|D|@]|E|$]]|$1|W|3|X|5|J|7|1E|8|@]|D|@]|E|$K|L]]|$1|Y|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|Z|$]]

This 3 workarounds uses <code>ascii_lowercase</code> from <code>string</code> module.
First, print while iterating:
<pre><code>from string import ascii_lowercase
text = &quot;You can have data without information, but you cannot have information without data.&quot;

for char in ascii_lowercase:
 count = 0
 for letter in text.lower():
 if letter == char:
 count += 1
 if count: 
 print(f'{char} : {count}')
</code></pre>
Second, using dictionary to update the counted value using <code>dict.update()</code>:
<pre><code>from string import ascii_lowercase
text = &quot;You can have data without information, but you cannot have information without data.&quot;

count = dict()
for char in ascii_lowercase:
 if char in text:
 count.update({char: text.lower().count(char)})
for char in count:
 print(f'{char} : {count[char]}')
</code></pre>
Third, using dict comprehension:
<pre><code>from string import ascii_lowercase
text = &quot;You can have data without information, but you cannot have information without data.&quot;
count = {char: text.lower().count(char) for char in ascii_lowercase if char in text}
for char in count: print(f'{char} : {count[char]}')
</code></pre>
Results:
<pre><code>a : 10
b : 1
c : 2
d : 2
e : 2
f : 2
h : 4
i : 6
m : 2
n : 7
o : 9
r : 2
t : 10
u : 5
v : 2
w : 2
y : 2
</code></pre>

I need to print out the number of a's, b's, c's etc. in the following statement and have it cascade down and have the number of times it appeared in the statement next to it.
&quot;you can have data without information, but you cannot have information without data.&quot;
I have got so far and could do with some help?
This is day 3 of me learning python so take it easy.
<pre><code>from collections import defaultdict

text = 'You can have data without information, but you cannot have information without data.'
lowerCaseText = text.lower()

print(lowerCaseText)
chars = defaultdict(int)

for char in text:
chars[char] += 1

print(chars)
</code></pre>
It needs to look like this
<pre><code>a : 10
b : 1
c : 2
d : 2
e : 2
f : 2
h : 4
i : 6
m : 2
n : 7
o : 9
r : 2
t : 10
u : 5
v : 2
w : 2
y : 2
</code></pre>

Python lists and and loops

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我需要在下面的语句中打印a's，b，c's等的数目，并将它级联下来，并在它旁边的语句中显示它的次数。“你可以拥有没有信息的数据，但没有数据你就不能拥有信息。”我到目前为止还能得到一些帮助吗？今天是我学习蟒蛇的第三天，所以别紧张。from collections import defaultdicttext = 'You...

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