blocks|key|2425469|text|在您的代码中，当您将文件内容转换为字典时，您将丢失顺序，因为dict是无序的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2425470|>>>+dict(csv.reader(db))
{
++'1':+'+#####+EASY+#####',
++'3':+'+#####+HARD+#####',
++'2':+'+#####+MEDIUM+#####',
++'key3':+'+value3',
++'key2':+'+value2',
++'key1':+'+value1',
++'key6':+'+value6',
++'key5':+'+value5',
++'key4':+'+value4'
}|code-block|syntax|javascript|2425471|例如，您可以迭代csv.reader(f)返回的内容，以准确映射文件中的内容：|2425472|mydict+=+{}

with+open("PLUG_dict.txt",+"r")+as+db:
++++for+key,+val+in+csv.reader(db):
++++++++if+len(key)+==+1:
+++++++++++#+Difficulty+line
+++++++++++dif+=+key
+++++++++++#+Construct+the+entry+for+this+difficulty,+and+save+its+name
+++++++++++mydict[dif]+=+{+'name'+:+val+}
++++++++else:
+++++++++++#+key,+val+line
+++++++++++#+Simply+add+it+in+the+dict+of+the+current+difficulty
+++++++++++mydict[dif][key]+=+val|2425473|如果文件的第一行不是“困难行”，这将不起作用。这将构造以下mydict|2425474|{
++'1':+{
++++'name':+'+#####+EASY+#####',
++++'key1':+'+value1',
++++'key2':+'+value2'
++},
++'3':+{
++++'name':+'+#####+HARD+#####',
++++'key5':+'+value5',
++++'key6':+'+value6'
++},
++'2':+{
++++'key3':+'+value3',
++++'name':+'+#####+MEDIUM+#####',
++++'key4':+'+value4'
++}
}|2425475|您确实可以更改处理'name'的方式。|2425476|然后，使用dif+=+'1'，您可以：|2425477|2425478|查找指定难度的值：|unordered-list-item|2425479|2425480|我的口令|2425481|将会返回|2425482|{+'name'：‘#####+EASY+#’，'key1'：‘value1'，'key2'：’value2‘}|2425483|查找难度下的元素个数：|2425484|2425485|len(mydictdif)+-1|2425486|2425487|2|2425488|在这里，-+1是从计数中删除'name'所必需的。根据您处理'name'的方式，它可以对字典中每个困难下的所有元素执行discarded.|2425489|Return操作：|2425490|2425491|{+key:对于key，mydict.values()中元素的val，如果key+!=+'name‘，则在elements.items()中使用val。}|2425492|2425493|{+'key3'：‘value3'，'key2'：’value2'，'key1'：‘value1'，'key6'：’value6'，'key5'：‘value5'，'key4'：’value4‘}|2425494|另外，根据您处理错误的方式，条件if+key+!=+'name'可以省略。|2425495|我不确定是否理解关于HARD错误的问题？|2425496|2425497|我希望它能有所帮助！|2425498|entityMap^0|U|4|0|0|8|D|0|0|T|6|0|0|9|6|0|5|9|0|0|0|0|0|0|0|0|0|0|0|0|4|3|E|6|U|6|0|0|0|0|0|0|G|G|0|A|4|0|0|0^^$0|@$1|2|3|4|5|6|7|1Y|8|@$9|1Z|A|20|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|21|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|22|8|@$9|23|A|24|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|25|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|26|8|@$9|27|A|28|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|29|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|2A|8|@$9|2B|A|2C|B|C]]|D|@]|E|$]]|$1|U|3|V|5|6|7|2D|8|@$9|2E|A|2F|B|C]]|D|@]|E|$]]|$1|W|3|-4|5|6|7|2G|8|@]|D|@]|E|$]]|$1|X|3|Y|5|Z|7|2H|8|@]|D|@]|E|$]]|$1|10|3|-4|5|6|7|2I|8|@]|D|@]|E|$]]|$1|11|3|12|5|6|7|2J|8|@]|D|@]|E|$]]|$1|13|3|14|5|6|7|2K|8|@]|D|@]|E|$]]|$1|15|3|16|5|6|7|2L|8|@]|D|@]|E|$]]|$1|17|3|18|5|Z|7|2M|8|@]|D|@]|E|$]]|$1|19|3|-4|5|6|7|2N|8|@]|D|@]|E|$]]|$1|1A|3|1B|5|6|7|2O|8|@]|D|@]|E|$]]|$1|1C|3|14|5|6|7|2P|8|@]|D|@]|E|$]]|$1|1D|3|1E|5|6|7|2Q|8|@]|D|@]|E|$]]|$1|1F|3|1G|5|6|7|2R|8|@$9|2S|A|2T|B|C]|$9|2U|A|2V|B|C]|$9|2W|A|2X|B|C]]|D|@]|E|$]]|$1|1H|3|1I|5|Z|7|2Y|8|@]|D|@]|E|$]]|$1|1J|3|-4|5|6|7|2Z|8|@]|D|@]|E|$]]|$1|1K|3|1L|5|6|7|30|8|@]|D|@]|E|$]]|$1|1M|3|14|5|6|7|31|8|@]|D|@]|E|$]]|$1|1N|3|1O|5|6|7|32|8|@]|D|@]|E|$]]|$1|1P|3|1Q|5|6|7|33|8|@$9|34|A|35|B|C]]|D|@]|E|$]]|$1|1R|3|1S|5|Z|7|36|8|@$9|37|A|38|B|C]]|D|@]|E|$]]|$1|1T|3|-4|5|6|7|39|8|@]|D|@]|E|$]]|$1|1U|3|1V|5|6|7|3A|8|@]|D|@]|E|$]]|$1|1W|3|-4|5|6|7|3B|8|@]|D|@]|E|$]]]|1X|$]]

In your code, when you transform the content of your file into a dictionary, you will lose the order, as <code>dict</code> is unordered:

<pre><code>&gt;&gt;&gt; dict(csv.reader(db))
{
 '1': ' ##### EASY #####',
 '3': ' ##### HARD #####',
 '2': ' ##### MEDIUM #####',
 'key3': ' value3',
 'key2': ' value2',
 'key1': ' value1',
 'key6': ' value6',
 'key5': ' value5',
 'key4': ' value4'
}
</code></pre>

You can for example iterate through what <code>csv.reader(f)</code> returns to map exactly what you have in your file:

<pre><code>mydict = {}

with open("PLUG_dict.txt", "r") as db:
 for key, val in csv.reader(db):
 if len(key) == 1:
 # Difficulty line
 dif = key
 # Construct the entry for this difficulty, and save its name
 mydict[dif] = { 'name' : val }
 else:
 # key, val line
 # Simply add it in the dict of the current difficulty
 mydict[dif][key] = val
</code></pre>

This will not work if the first line of your file is not a "Difficulty line".
This will construct the following <code>mydict</code>

<pre><code>{
 '1': {
 'name': ' ##### EASY #####',
 'key1': ' value1',
 'key2': ' value2'
 },
 '3': {
 'name': ' ##### HARD #####',
 'key5': ' value5',
 'key6': ' value6'
 },
 '2': {
 'key3': ' value3',
 'name': ' ##### MEDIUM #####',
 'key4': ' value4'
 }
}
</code></pre>

You can indeed change the way <code>'name'</code> is handled.

Then, with <code>dif = '1'</code>, you can:

<ul>
<li>Find the values for a specified difficulty:

<pre><code>mydict[dif]
</code></pre>

will return

<pre><code>{
 'name': ' ##### EASY #####',
 'key1': ' value1',
 'key2': ' value2'
}
</code></pre></li>
<li>Find the number of element under a difficulty:

<pre><code>len(mydict[dif]) - 1
</code></pre>

will return

<pre><code>2
</code></pre>

Here, the <code>- 1</code> is necessary to remove <code>'name'</code> from the count. Depending on how you handle <code>'name'</code>, it could be discarded.</li>
<li>Return all elements under every difficulty in a dictionary:

<pre><code>{
 key: val
 for elements in mydict.values()
 for key, val in elements.items()
 if key != 'name'
}
</code></pre>

will return

<pre><code>{
 'key3': ' value3',
 'key2': ' value2',
 'key1': ' value1',
 'key6': ' value6',
 'key5': ' value5',
 'key4': ' value4'
}
</code></pre>

Also here, depending on how you handle <code>'name'</code>, the condition <code>if key != 'name'</code> can be omitted.</li>
<li>I am not sure to understand the question about <code>HARD</code> error?</li>
</ul>

I hope it helps!

blocks|key|2425526|text|你可以按照下面的思路做一些事情：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2425527|import+csv

with+open('dictionary.txt',+'r')+as+f:
++++data={}
++++k=None+
++++for+row+in+csv.reader(f):
++++++++print+row
++++++++if+row[1]+in+'+#####+EASY+#####+#####+MEDIUM+#####+#####+HARD+#####':
++++++++++++k=row[1].strip()
++++++++++++data[k]={}
++++++++elif+k:
++++++++++++data[k][row[0]]=row[1]|code-block|syntax|javascript|2425528|产生：|2425529|>>>+data
{'#####+HARD+#####':+{'key6':+'+value6',+'key5':+'+value5'},+'#####+EASY+#####':+{'key2':+'+value2',+'key1':+'+value1'},+'#####+MEDIUM+#####':+{'key3':+'+value3',+'key4':+'+value4'}}|2425530|然后使用data[category][key_you_want_in_category]进行访问|offset|length|style|CODE|2425531|entityMap^0|0|0|0|0|4|14|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

You can do something along these lines:

<pre><code>import csv

with open('dictionary.txt', 'r') as f:
 data={}
 k=None 
 for row in csv.reader(f):
 print row
 if row[1] in ' ##### EASY ##### ##### MEDIUM ##### ##### HARD #####':
 k=row[1].strip()
 data[k]={}
 elif k:
 data[k][row[0]]=row[1]
</code></pre>

Produces:

<pre><code>&gt;&gt;&gt; data
{'##### HARD #####': {'key6': ' value6', 'key5': ' value5'}, '##### EASY #####': {'key2': ' value2', 'key1': ' value1'}, '##### MEDIUM #####': {'key3': ' value3', 'key4': ' value4'}}
</code></pre>

Then access with <code>data[category][key_you_want_in_category]</code>

blocks|key|2425560|text|请注意，Python的数据类型比csv更适合这种情况|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2425561|改为使用plug_dict.py，并将此代码放入|2425562|levels+=+{'EASY':+["value+1",+"value+2"],
++++++'MEDIUM':+["value+3",+"value+4"],
++++++'HARD':+["value+5",+"value+5"]
}|code-block|syntax|javascript|2425563|并做到：|2425564|from+plug_dict.py+import+levels

SECTIONS+=+[]


#index+of+a+particular+line
Unnecessary,+get+the+values+instead
values_of_sect+=+levels['HARD']

#+THE+DIFFICULTIES
for+key+in+levels:
++++SECTIONS.append(key)


#+no+of+elements+in+between+difficulties
#+do
elements+=+len(levels['HARD'])

#+Handle+the+'HARD'+difficulty
unnecessary,+python+types+handles+it+beautifully,+you+will+get+no+errors|2425565|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Please Python's Data types are more suitable for this than the csv

use plug_dict.py instead and put this inside

<pre><code>levels = {'EASY': ["value 1", "value 2"],
 'MEDIUM': ["value 3", "value 4"],
 'HARD': ["value 5", "value 5"]
}
</code></pre>

and do:

<pre><code>from plug_dict.py import levels

SECTIONS = []


#index of a particular line
Unnecessary, get the values instead
values_of_sect = levels['HARD']

# THE DIFFICULTIES
for key in levels:
 SECTIONS.append(key)


# no of elements in between difficulties
# do
elements = len(levels['HARD'])

# Handle the 'HARD' difficulty
unnecessary, python types handles it beautifully, you will get no errors
</code></pre>

I have a CSV dictionary dictionary.txt that I want to use in my Python program. I already use the standard library csv to open the document. These are levels of difficulty the user will input to select a difficulty. If the user selects "EASY", I want to be able to ignore the headings, and be able to get the number of elements under the heading "EASY", up to just before the next difficulty. The number before the header is the level of difficulty. I have headings and data between those headings, similar to:

<pre><code>1, ##### EASY #####
key1, value1
key2, value2
2, ##### MEDIUM #####
key3, value3
key4, value4
3, ##### HARD #####
key5, value5
key6, value6
</code></pre>

I want to return a dictionary-type of the useful elements in the dictionary (just key n, value n) for each section, and also an "ALL" option. My current code looks like this, but I don't know what to do from here:

<pre><code>with open("PLUG_dict.txt", "r") as db:
 mydict = dict(csv.reader(db))
for key in mydict.keys():
 if len(key) == 1:
 sections.append(mydict[key])
</code></pre>

How do I: 

<ul>
<li>Find the index of a line of a specified difficulty (beginning and end)?</li>
<li>Find the number of elements in between difficulties?</li>
<li>Handle the last difficulty, "HARD", such that it won't become an error to run it to the end of the dictionary?</li>
<li>If I wanted to include a classification, "ALL", that returned just all the useful elements in the dictionary, how to do that? </li>
</ul>

Python CSV Parsing Dictionary and Finding Elements' Indices

Python

我有一个CSV字典dictionary.txt，我想在我的Python程序中使用它。我已经使用标准库csv打开了文档。这些是用户将输入以选择难度的难度级别。如果用户选择"EASY"，我希望能够忽略标题，并能够获得“EASY”标题下的元素数量，直到下一个难度之前。标题前的数字是难度级别。我有标题和这些标题之间的数据，类似...

问Python CSV解析字典和元素索引的查找
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