blocks|key|1352376|text|如果需要的话，用字典创建一个列表，然后用df+=+pd.DataFrame.from_records(your_list)创建一个新的数据格式。List的“追加”方法非常有效，不会被废弃。另一方面，由于数据的设计，常常需要重新创建数据，并将所有数据复制到附件中--这就是为什么它们不推荐使用这种方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1352377|entityMap^0|K|15|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

Create a list with your dictionaries, if they are needed, and then create a new dataframe with <code>df = pd.DataFrame.from_records(your_list)</code>. List's &quot;append&quot; method are very efficient and won't be ever deprecated. Dataframes on the other hand, frequently have to be recreated and all data copied over on appends, due to their design - that is why they deprecated the method

blocks|key|2784644|text|我也喜欢附加方法。但是你可以用一串数据来做这件事|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2784645|df+=+pd.concat([df,+pd.DataFrame.from_records([{+'a':+1,+'b':+2+}])])|code-block|syntax|javascript|2784646|或对具有增量上升索引的DataFrames上的值使用loc和元组。|2784647|df.loc[len(df),+['a','b']]+=+1,+2|2784648|或者也许|2784649|df.loc[len(df),+df.columns]+=+3,+4|2784650|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|$]]

I also like the append method.
But you can do it in one line with a list of dicts
<pre><code>df = pd.concat([df, pd.DataFrame.from_records([{ 'a': 1, 'b': 2 }])])
</code></pre>
or using loc and tuples for values on DataFrames with incremenal ascending indexes
<pre><code>df.loc[len(df), ['a','b']] = 1, 2
</code></pre>
or maybe
<pre><code>df.loc[len(df), df.columns] = 3, 4
</code></pre>

blocks|key|2784667|text|如果您想要使用concat，那么：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2784668|append|2784669|outputxlsx+=+outputxlsx.append(df,+ignore_index=True)|code-block|syntax|javascript|2784670|concat|2784671|outputxlsx+=+pd.concat([outputxlsx,+df])|2784672|entityMap^0|7|6|0|0|6|0|0|0|6|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

If you want to use <code>concat</code> instead:
<code>append</code>
<pre class="lang-py prettyprint-override"><code>outputxlsx = outputxlsx.append(df, ignore_index=True)
</code></pre>
<code>concat</code>
<pre class="lang-py prettyprint-override"><code>outputxlsx = pd.concat([outputxlsx, df])
</code></pre>

blocks|key|1352431|text|我也面临着一个类似的问题。其他的解决方案对我并不有效。我把这个答案留在这里，作为一个额外的可能性来处理这个问题，因为这是google对某些搜索的第一个结果，而我本人至少第二次在这里结束了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1352432|在我的例子中，数据不是dict，而是一组已知参数的值列表。我希望将参数值作为行添加到dataframe中，因为这样我可以通过of参数访问一个参数的一系列所有值。|1352433|我从一个空的DataFrame开始：|1352434|parameters+=+['a',+'b',+'c',+'d',+'e',+'f']
df+=+pd.DataFrame(columns=parameters)|code-block|syntax|javascript|1352435|df：|1352436|++++++++a+++b+++c+++d+++e+++f|1352437|通过追加，我可以添加非常方便的行，如下所示：|1352438|new_row+=+pd.Series([1,2,3,4,5,6],+index=parameters,+name='row1')
df.append(new_row)|1352439|1352440|++++++++a+++b+++c+++d+++e+++f
row1++++1+++2+++3+++4+++5+++6|1352441|使用pd.concat，我发现它以非常类似的方式提供了相同的结果：|1352442|new_row+=+pd.DataFrame([1,2,3,4,5,6],+columns=['row1'],+index=parameters).T
df+=+pd.concat((df,+new_row))|1352443|关键是从1d数据创建一个(1，n)数据，然后转换它以匹配其他数据。|1352444|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|14|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|15|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|16|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|17|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|18|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|19|8|@]|9|@]|A|$]]|$1|Q|3|R|5|H|7|1A|8|@]|9|@]|A|$I|J]]|$1|S|3|L|5|6|7|1B|8|@]|9|@]|A|$]]|$1|T|3|U|5|H|7|1C|8|@]|9|@]|A|$I|J]]|$1|V|3|W|5|6|7|1D|8|@]|9|@]|A|$]]|$1|X|3|Y|5|H|7|1E|8|@]|9|@]|A|$I|J]]|$1|Z|3|10|5|6|7|1F|8|@]|9|@]|A|$]]|$1|11|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|12|$]]

I was facing a similar issue. The other solutions weren't really working for me. I'm leaving this answer here as an additional possibility to deal with the issue since this is the first google result for certain searches and I myself ended here at least for the second time.
In my case the data is not a dict but just a list of values for a known set of parameters.
I want to add the parameter values to a dataframe as rows because this way I can access a series of all the values for one parameter via df[parameter].
I start with an empty DataFrame:
<pre><code>parameters = ['a', 'b', 'c', 'd', 'e', 'f']
df = pd.DataFrame(columns=parameters)
</code></pre>
df:
<pre><code> a b c d e f
</code></pre>
With append I could add rows very convenient like so:
<pre><code>new_row = pd.Series([1,2,3,4,5,6], index=parameters, name='row1')
df.append(new_row)
</code></pre>
df:
<pre><code> a b c d e f
row1 1 2 3 4 5 6
</code></pre>
With pd.concat I found this to deliver the same result in very similar way:
<pre><code>new_row = pd.DataFrame([1,2,3,4,5,6], columns=['row1'], index=parameters).T
df = pd.concat((df, new_row))
</code></pre>
The key was to create a (1,n) dataframe from the 1d data and then transpose it to match the other dataframe.

blocks|key|2476293|text|对于那些想要一个描述函数而不是很多一行的人，像我这样的人，这里有一个基于@Rafael的选项。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2476294|def+appendDictToDF(df,dictToAppend):
++df+=+pd.concat([df,+pd.DataFrame.from_records([dictToAppend])])
++return+df

#+Creating+an+empty+dataframe
df+=+pd.DataFrame(columns=['a',+'b'])

#+Appending+a+row
df=+appendDictToDF(df,{+'a':+1,+'b':+2+})|code-block|syntax|javascript|2476295|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|$]]

For those, like me, who want a descriptive function rather than lots of one-liners, here is an option based on @Rafael Gaitan above.
<pre><code>def appendDictToDF(df,dictToAppend):
 df = pd.concat([df, pd.DataFrame.from_records([dictToAppend])])
 return df

# Creating an empty dataframe
df = pd.DataFrame(columns=['a', 'b'])

# Appending a row
df= appendDictToDF(df,{ 'a': 1, 'b': 2 })
</code></pre>

blocks|key|2476247|text|#+Deprecated+issue+has+been+resolved

#+Creating+an+empty+dataframe
df+=+pd.DataFrame(columns=['a',+'b'])
print("df+columns:",+df)

#+Appending+a+row
df+=+df.append({+'a':+1,+'b':+2+},+ignore_index=True)
print("df+column+Values+:",+df)

#+Create+the+new+row+as+its+own+dataframe
df_new_row+=+pd.DataFrame.from_records({+'a':+[3],+'b':+[4]+})
df+=+pd.concat([df,+df_new_row])
print("pd+concat+with+two+df's+:",+df)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2476248|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code># Deprecated issue has been resolved

# Creating an empty dataframe
df = pd.DataFrame(columns=['a', 'b'])
print(&quot;df columns:&quot;, df)

# Appending a row
df = df.append({ 'a': 1, 'b': 2 }, ignore_index=True)
print(&quot;df column Values :&quot;, df)

# Create the new row as its own dataframe
df_new_row = pd.DataFrame.from_records({ 'a': [3], 'b': [4] })
df = pd.concat([df, df_new_row])
print(&quot;pd concat with two df's :&quot;, df)
</code></pre>

blocks|key|2784674|text|我以前在我的程序中使用DataFrame.append时也有这个问题，但现在已经修复了。希望这个片段能帮上忙。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2784675|import+pandas+as+pd
df1=pd.DataFrame(dict_1)

def+addData(param1,param2,param3):
++++dict_2={"list1":var1,+"list2":var2,+"list3":var3}
++++df2=pdDataFrame(dict_2,+index={len(dict_2)%2B1})
++++dfc=pd.concat([df1,+df2])
++++return+dfc|code-block|syntax|javascript|2784676|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|$]]

i also have the problem when using DataFrame.append in my program before, but it has been fixed now. Hopefully this snippet can help so.
<pre><code>import pandas as pd
df1=pd.DataFrame(dict_1)

def addData(param1,param2,param3):
 dict_2={&quot;list1&quot;:var1, &quot;list2&quot;:var2, &quot;list3&quot;:var3}
 df2=pdDataFrame(dict_2, index={len(dict_2)+1})
 dfc=pd.concat([df1, df2])
 return dfc
</code></pre>

I use the following method a lot to append a single row to a dataframe. One thing I really like about it is that it allows you to append a simple dict object. For example:
<pre><code># Creating an empty dataframe
df = pd.DataFrame(columns=['a', 'b'])

# Appending a row
df = df.append({ 'a': 1, 'b': 2 }, ignore_index=True)
</code></pre>
Again, what I like most about this is that the code is very clean and requires very few lines. Now I suppose the recommended alternative is:
<pre><code># Create the new row as its own dataframe
df_new_row = pd.DataFrame({ 'a': [1], 'b': [2] })
df = pd.concat([df, df_new_row])
</code></pre>
So what was one line of code before is now two lines with a throwaway variable and extra cruft where I create the new dataframe. :( Is there a good way to do this that just uses a dict like I have in the past (that is not deprecated)?

Good alternative to Pandas .append() method, now that it is being deprecated?

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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我经常使用下面的方法将一行附加到dataframe。我真正喜欢它的一件事是，它允许您附加一个简单的dict对象。例如：# Creating an empty dataframedf = pd.DataFrame(columns=['a', 'b'])# Appending a rowdf = df.append({ 'a': 1, 'b': 2 }, ignore_index=True)同样，我最

问Pandas .append()方法的好替代品，现在它已经被废弃了吗？
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问Pandas .append()方法的好替代品，现在它已经被废弃了吗？EN

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问Pandas .append()方法的好替代品，现在它已经被废弃了吗？
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