entityMap|blocks|key|3o23j|text|我认为value+=+operationData.iloc[[row_indexer],[iloc_indexer]]返回了一个数据帧。试试value+=+value.loc[col].values[0]。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4od8f^0|3|1K|1Y|W|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@$A|I|B|J|C|D]|$A|K|B|L|C|D]]|E|@]|F|$]]|$2|G|4|-4|6|7|8|M|9|@]|E|@]|F|$]]]]

I think <code>value = operationData.iloc[[row_indexer],[iloc_indexer]]</code> returns a dataframe. Try <code>value = value.loc[col].values[0]</code>.

entityMap|blocks|key|2dlbj|text|我不确定这是否会确切地帮助你，但也许它会让你走上正确的方向。您可以使用Pandas.DataFrame.itertuples()遍历数据帧中的所有行，在需要的时候摘取值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|ib6c|我更进一步，创建了一个快速的列标签字典来帮助同步嵌套循环。|7sfau|我试着在必要的地方进行评论，但是如果你不理解什么，请让我知道！|5hf1c|import+pandas+as+pd
import+openpyxl
wb+=+load_workbook(filename='./generic_workbook_name.xlsx')

#+Created+smoe+data+for+a+dataframe.
fake_data_dict+=+{
++++'OpNo':['1','2','3','4',],
++++'StationNo':['11','22','33','44',],
++++'Spindle':['S1','S2','S3','S4',],
++++'OpDescription':['This','is','a','description',]
++++}

#+Create+the+dataframe.
data+=+pd.DataFrame(fake_data_dict)|code-block|syntax|javascript|9a09d|我们的数据框架：|68ej9|++OpNo+StationNo+Spindle+OpDescription
0++++1++++++++11++++++S1++++++++++This
1++++2++++++++22++++++S2++++++++++++is
2++++3++++++++33++++++S3+++++++++++++a
3++++4++++++++44++++++S4+++description|5j4h3|脚本的其余部分：|bnl1h|col_list+=+['OpNo','StationNo','Spindle','OpDescription']


#+Create+a+column+label+dictionary;+Add+1+to+index+for+Excel+cells
col_dict+=+{i%2B1:v+for+i,+v+in+enumerate(col_list)}

#+Iterate+over+each+row
for+idx,+row+in+enumerate(data.itertuples(),+start+=+1):
++++#+For+each+key+in+our+column+dictionary+[0,+1,+2,+3]
++++for+key+in+col_dict.keys():
++++++++print('Row:+{a}\n\tColumn:+{b}\n\t\tValue:+{c}'.format(a+=+idx,+b+=+key,
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++#+Reduce+the+index+by+1;+Get+column+name+based+on+key+value.
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++c+=+data.loc[idx+-+1,+col_dict[key]]))|5uugb|输出：|786k9|Row:+1
++++Column:+1
++++++++Value:+1
Row:+1
++++Column:+2
++++++++Value:+11
Row:+1
++++Column:+3
++++++++Value:+S1
Row:+1
++++Column:+4
++++++++Value:+This
Row:+2
++++Column:+1
++++++++Value:+2
Row:+2
++++Column:+2
++++++++Value:+22
Row:+2
++++Column:+3
++++++++Value:+S2
Row:+2
++++Column:+4
++++++++Value:+is
Row:+3
++++Column:+1
++++++++Value:+3
Row:+3
++++Column:+2
++++++++Value:+33
Row:+3
++++Column:+3
++++++++Value:+S3
Row:+3
++++Column:+4
++++++++Value:+a
Row:+4
++++Column:+1
++++++++Value:+4
Row:+4
++++Column:+2
++++++++Value:+44
Row:+4
++++Column:+3
++++++++Value:+S4
Row:+4
++++Column:+4
++++++++Value:+description|1emsi|考虑到这一点，这可以简化您的脚本：|frtl3|for+idx,+row+in+enumerate(data.itertuples(),+start+=+1):
++++for+key+in+col_dict.keys():
++++++++wb['OneOpSheet'].cell(row+=+(idx+%2B+11),+column+=+(key+%2B+1)).value+=+data.loc[idx+-+1,+col_dict[key]]|70g4n^0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|12|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|13|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|14|9|@]|A|@]|B|$]]|$2|G|4|H|6|I|8|15|9|@]|A|@]|B|$J|K]]|$2|L|4|M|6|7|8|16|9|@]|A|@]|B|$]]|$2|N|4|O|6|I|8|17|9|@]|A|@]|B|$J|K]]|$2|P|4|Q|6|7|8|18|9|@]|A|@]|B|$]]|$2|R|4|S|6|I|8|19|9|@]|A|@]|B|$J|K]]|$2|T|4|U|6|7|8|1A|9|@]|A|@]|B|$]]|$2|V|4|W|6|I|8|1B|9|@]|A|@]|B|$J|K]]|$2|X|4|Y|6|7|8|1C|9|@]|A|@]|B|$]]|$2|Z|4|10|6|I|8|1D|9|@]|A|@]|B|$J|K]]|$2|11|4|-4|6|7|8|1E|9|@]|A|@]|B|$]]]]

I'm not sure if this is going to help you exactly, but maybe it'll get you in the right direction. You can use Pandas.DataFrame.itertuples() to run across all rows in your dataframe, picking off values as you need them.

I went a bit further and created a quick column label dictionary to help sync the nested loops.

I tried to comment where necessary, but if you don't understand something, let me know!

<pre><code>import pandas as pd
import openpyxl
wb = load_workbook(filename='./generic_workbook_name.xlsx')

# Created smoe data for a dataframe.
fake_data_dict = {
 'OpNo':['1','2','3','4',],
 'StationNo':['11','22','33','44',],
 'Spindle':['S1','S2','S3','S4',],
 'OpDescription':['This','is','a','description',]
 }

# Create the dataframe.
data = pd.DataFrame(fake_data_dict)
</code></pre>

Our dataframe:

<pre><code> OpNo StationNo Spindle OpDescription
0 1 11 S1 This
1 2 22 S2 is
2 3 33 S3 a
3 4 44 S4 description
</code></pre>

The rest of the script:

<pre><code>col_list = ['OpNo','StationNo','Spindle','OpDescription']


# Create a column label dictionary; Add 1 to index for Excel cells
col_dict = {i+1:v for i, v in enumerate(col_list)}

# Iterate over each row
for idx, row in enumerate(data.itertuples(), start = 1):
 # For each key in our column dictionary [0, 1, 2, 3]
 for key in col_dict.keys():
 print('Row: {a}\n\tColumn: {b}\n\t\tValue: {c}'.format(a = idx, b = key,
 # Reduce the index by 1; Get column name based on key value.
 c = data.loc[idx - 1, col_dict[key]]))
</code></pre>

Output:

<pre><code>Row: 1
 Column: 1
 Value: 1
Row: 1
 Column: 2
 Value: 11
Row: 1
 Column: 3
 Value: S1
Row: 1
 Column: 4
 Value: This
Row: 2
 Column: 1
 Value: 2
Row: 2
 Column: 2
 Value: 22
Row: 2
 Column: 3
 Value: S2
Row: 2
 Column: 4
 Value: is
Row: 3
 Column: 1
 Value: 3
Row: 3
 Column: 2
 Value: 33
Row: 3
 Column: 3
 Value: S3
Row: 3
 Column: 4
 Value: a
Row: 4
 Column: 1
 Value: 4
Row: 4
 Column: 2
 Value: 44
Row: 4
 Column: 3
 Value: S4
Row: 4
 Column: 4
 Value: description
</code></pre>

With that in mind, this could simplify your script:

<pre><code>for idx, row in enumerate(data.itertuples(), start = 1):
 for key in col_dict.keys():
 wb['OneOpSheet'].cell(row = (idx + 11), column = (key + 1)).value = data.loc[idx - 1, col_dict[key]]
</code></pre>

entityMap|blocks|key|chjtu|text|我需要将这两个命令都放在for循环中，并在完成后重置两个索引器。下面的嵌套循环完成了我所需要的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4qkse|while+row_indexer+<+count_row:++++
++++for+col+in+col_list:
++++++++value+=+operationData.iloc[[row_indexer],[iloc_indexer]]
++++++++value+=+value[col].values[0]
++++++++wb['OneOpSheet'].cell(row+=+(row_indexer+%2B+12),+column+=+(col_indexer+%2B+2)).value+=+value
++++++++col_indexer+%2B=+1
++++++++iloc_indexer+%2B=+1
++++row_indexer+%2B=+1
++++iloc_indexer+=+0
++++col_indexer+=+0|code-block|syntax|javascript|dgpp7^0|C|3|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@$A|N|B|O|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|P|9|@]|E|@]|F|$J|K]]|$2|L|4|-4|6|7|8|Q|9|@]|E|@]|F|$]]]]

I needed to have both commands inside the <code>for</code> loop and reset a couple of the indexers after that finished. The below nested loop accomplishes what I needed:

<pre><code>while row_indexer &lt; count_row: 
 for col in col_list:
 value = operationData.iloc[[row_indexer],[iloc_indexer]]
 value = value[col].values[0]
 wb['OneOpSheet'].cell(row = (row_indexer + 12), column = (col_indexer + 2)).value = value
 col_indexer += 1
 iloc_indexer += 1
 row_indexer += 1
 iloc_indexer = 0
 col_indexer = 0
</code></pre>

I have a nested loop that I'm trying to pass values from a list into, but it will not recognize the list value. If I replace the <code>value[col]</code> with any list value like <code>value['OpNo']</code> it works. Is there a specific wrapper or something I need around either the list values or the assignment code?

I tried the lists like <code>col_list = ["'OpNo'", "'StationNo'", "'Spindle'", "'OpDescription'"]</code> and wrapping the <code>col</code> callout like <code>value[[col]]</code>

The code below is getting the following exception: <code>KeyError: 'OpNo'</code>

<pre><code>row_indexer = 0
col_indexer = 1
iloc_indexer = 0
count_row = operationData.shape[0]
col_list = ['OpNo', 'StationNo', 'Spindle', 'OpDescription']
while row_indexer &lt; count_row: 
 value = operationData.iloc[[row_indexer],[iloc_indexer]]
 for col in col_list:
 value = value[col].values[0]
 wb['OneOpSheet'].cell(row = (row_indexer + 12), column = (col_indexer + 1)).value = value
 col_indexer = (col_indexer + 1)
 row_indexer = (row_indexer + 1)
 iloc_indexer = (iloc_indexer + 1)
</code></pre>

Pandas - how to pass variable as column into nested loop?

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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 我有一个嵌套循环，我试图将列表中的值传递到该循环中，但它无法识别列表值。如果我用value['OpNo']之类的任何列表值替换value[col]，它就可以工作。对于列表值或赋值代码，是否有特定的包装器或我需要的东西？ 我尝试了像col_list = ["'OpNo'", "'StationNo'", "'Spind...

问Pandas -如何将变量作为列传递到嵌套循环中？
EN

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问Pandas -如何将变量作为列传递到嵌套循环中？EN