blocks|key|281524|text|这个错误的发生是因为熊猫需要一个指数。一开始，这似乎有点令人困惑，因为您会想到列表索引。这实际上要求的是每个字典对应于每个字典的列号。您可以这样设置：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|281525|import+pandas+as+pd
list+=+['a',+'b',+'c',+'d']
df+=+pd.DataFrame(list,+index+=+[0,+1,+2,+3])|code-block|syntax|javascript|281526|然后，数据框架产生：|281527|+++0++
0+'a'
1+'b'
2+'c'
3+'d'|281528|特别是对于您来说，使用numpy+(未测试)可以这样做：|281529|list_of_dfs+=+{}+

for+I+in+range(0,len(regionLoadArray)):
++++list_of_dfs[I]+=+pd.read_csv(regionLoadArray[I])+

ind+=+np.arange[len(list_of_dfs)]

dataframe+=+pd.DataFrame(list_of_dfs,+index+=+ind)|281530|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

This error occurs because pandas needs an index. At first this seems sort of confusing because you think of list indexing. What this is essentially asking for is a column number for each dictionary to correspond to each dictionary. You can set this like so:

<pre><code>import pandas as pd
list = ['a', 'b', 'c', 'd']
df = pd.DataFrame(list, index = [0, 1, 2, 3])
</code></pre>

The data frame then yields:

<pre><code> 0 
0 'a'
1 'b'
2 'c'
3 'd'
</code></pre>

For you specifically, this might look something like this using numpy (not tested):

<pre><code>list_of_dfs = {} 

for I in range(0,len(regionLoadArray)):
 list_of_dfs[I] = pd.read_csv(regionLoadArray[I]) 

ind = np.arange[len(list_of_dfs)]

dataframe = pd.DataFrame(list_of_dfs, index = ind)
</code></pre>

blocks|key|3028377|text|不幸的是，熊猫在创建DataFrame时总是需要一个索引。您可以自己设置它，也可以使用具有以下结构的对象，这样熊猫就可以确定索引本身：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3028378|++++data=+{'a':[1],'b':[2]}|code-block|syntax|javascript|3028379|因为在你的情况下编辑数据并不容易，|3028380|hacky解决方案是将数据打包到列表中。|3028381|++++dataframe+=+pd.DataFrame([list_of_dfs])|3028382|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Pandas unfortunately always needs an index when creating a DataFrame.
You can either set it yourself, or use an object with the following structure so pandas can determine the index itself:
<pre><code> data= {'a':[1],'b':[2]}
</code></pre>
Since it won't be easy to edit the data in your case,
A hacky solution is to wrap the data into a list
<pre><code> dataframe = pd.DataFrame([list_of_dfs])
</code></pre>

blocks|key|3028404|text|根据您的需要，一个简单的解决方法可以是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3028405|dct+=+{'col1':+'abc',+'col2':+123}
dct+=+{k:[v]+for+k,v+in+dct.items()}++#+WORKAROUND
df+=+pd.DataFrame(dct)|code-block|syntax|javascript|3028406|这会导致|3028407|print(df)

++col1++col2
0++abc+++123|3028408|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Depending on your needs, a simple workaround could be:
<pre><code>dct = {'col1': 'abc', 'col2': 123}
dct = {k:[v] for k,v in dct.items()} # WORKAROUND
df = pd.DataFrame(dct)
</code></pre>
which results in
<pre><code>print(df)

 col1 col2
0 abc 123
</code></pre>

blocks|key|281614|text|import+pandas+as+pd

d+=+[{"a":+1,+"b":2,+"c":+3},
++++{"a":+4,+"b":5,+"c":+6},
++++{"a":+7,+"b":8,+"c":+9}
]

pd.DataFrame(d,+index=list(range(len(d))))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|281615|返回：|unstyled|281616|+++a++++b+++c
0+++1+++2+++3
1+++4+++5+++6
2+++7+++8+++9|281617|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>import pandas as pd

d = [{&quot;a&quot;: 1, &quot;b&quot;:2, &quot;c&quot;: 3},
 {&quot;a&quot;: 4, &quot;b&quot;:5, &quot;c&quot;: 6},
 {&quot;a&quot;: 7, &quot;b&quot;:8, &quot;c&quot;: 9}
]

pd.DataFrame(d, index=list(range(len(d))))
</code></pre>
returns:
<pre><code> a b c
0 1 2 3
1 4 5 6
2 7 8 9
</code></pre>

Currently, I am using a for loop to read csv files from a folder.
After reading the csv file, I am storing the data into one row of a dictionary.
When I print the data types using "print(list_of_dfs.dtypes)" I receive:

dtype: object
DATETIME : object
VALUE : float64
ID : int64
ID Name: object.

Note that this is a nested dictionary with thousands of values stored in each of these data fields. I have 26 rows of the structure listed above. I am trying to append the dictionary rows into a dataframe where I will have only 1 row consisting of the datafields:

Index DATETIME VALUE ID ID Name.

Note: I am learning python as I go.
I tried using an array to store the data and then convert the array to a dataframe but I could not append the rows of the dataframe.

Using the dictionary method I attempted "df = pd.Dataframe(list_of_dfs)"
This throws an error.

<pre><code>list_of_dfs = {} 

for I in range(0,len(regionLoadArray)
list_of_dfs[I] = pd.read_csv(regionLoadArray[I]) 

#regionLoadArray contains my- file names from list directory.

dataframe = pd.DataFrame(list_of_dfs)
#this method was suggested at thispoint.com for nested dictionaries.
#This is where my error occurs^
</code></pre>

<blockquote>
 ValueError: If using all scalar values, you must pass an index
</blockquote>

I appreciate any assistance with this issue as I am new to python.
My current goals is to simply produce a dataframe with my Headers that I can then send to a csv.

Dictionary to Dataframe Error: "If using all scalar values, you must pass an index"

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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目前，我正在使用for循环从文件夹中读取csv文件。读取csv文件后，我将数据存储在字典的一行中。当我使用" print (list_of_dfs.dtypes)“打印数据类型时，我会收到：类型:对象日期时间:对象值: float64 ID : int64 ID名称: object。请注意，这是一个嵌套字典，每个数据字...

问对Dataframe错误的字典：“如果使用所有标量值，则必须传递索引”
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问对Dataframe错误的字典：“如果使用所有标量值，则必须传递索引”EN