blocks|key|1934368|text|熊猫痛苦地工作在列表中，这里有一些棘手的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1934369|s+=+s.fillna(pd.Series([[0,0,0]]+*+len(s),+index=s.index))
print+(s)
0++++[1,+2,+3]
1++++[1,+2,+3]
2++++[0,+0,+0]
3++++[1,+2,+3]
4++++[1,+2,+3]
5++++[0,+0,+0]
dtype:+object|code-block|syntax|javascript|1934370|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|$]]

Pandas working with list with pain, here is hacky solution:
<pre><code>s = s.fillna(pd.Series([[0,0,0]] * len(s), index=s.index))
print (s)
0 [1, 2, 3]
1 [1, 2, 3]
2 [0, 0, 0]
3 [1, 2, 3]
4 [1, 2, 3]
5 [0, 0, 0]
dtype: object
</code></pre>

blocks|key|1068825|text|Series.reindex|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1068826|s.dropna().reindex(s.index,+fill_value=[0,+0,+0])|code-block|syntax|javascript|1068827|0++++[1,+2,+3]
1++++[1,+2,+3]
2++++[0,+0,+0]
3++++[1,+2,+3]
4++++[1,+2,+3]
5++++[0,+0,+0]
dtype:+object|1068828|entityMap^0|0|E|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|H|7|S|8|@]|D|@]|E|$I|J]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

<h2><code>Series.reindex</code></h2>
<pre><code>s.dropna().reindex(s.index, fill_value=[0, 0, 0])
</code></pre>
<hr />
<pre><code>0 [1, 2, 3]
1 [1, 2, 3]
2 [0, 0, 0]
3 [1, 2, 3]
4 [1, 2, 3]
5 [0, 0, 0]
dtype: object
</code></pre>

blocks|key|1068954|text|文档表示此值不能是list。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1068955|用于填充孔的值(例如0)，交替地是一个dict/Series/DataFrame值，该值指定每个索引(对一个序列)或列(对于一个DataFrame)使用哪个值。未在dict/Series/DataFrame中填充的值将不会被填充。这个值不能是一个列表.|blockquote|BOLD|1068956|这可能是当前实现的一个限制，如果没有修补源代码，您就必须求助于解决办法(如下所示)。|1068957|但是，如果您不打算使用交错数组，那么您真正想要做的可能是用pd.Series()替换pd.DataFrame()，例如：|1068958|import+numpy+as+np
import+pandas+as+pd


s+=+pd.DataFrame(
++++++++[[1,+2,+3],
+++++++++[1,+2,+3],
+++++++++[np.nan],
+++++++++[1,+2,+3],
+++++++++[1,+2,+3],
+++++++++[np.nan]],
++++++++dtype=pd.Int64Dtype())++#+to+mix+integers+with+NaNs


s.fillna(0)
#++++0++1++2
#+0++1++2++3
#+1++1++2++3
#+2++0++0++0
#+3++1++2++3
#+4++1++2++3
#+5++0++0++0|code-block|syntax|javascript|1068959|如果您确实需要使用锯齿数组，您可以使用其他答案中的任何一种建议解决方法，或者可以使您的尝试之一发挥作用，例如：|1068960|ii+=+s.isna()
nn+=+ii.sum()
s[ii]+=+pd.Series([[0,+0,+0]]+*+nn).to_numpy()
#+0++++[1,+2,+3]
#+1++++[1,+2,+3]
#+2++++[0,+0,+0]
#+3++++[1,+2,+3]
#+4++++[1,+2,+3]
#+5++++[0,+0,+0]
#+dtype:+object|1068961|它基本上使用NumPy掩蔽来填充系列。诀窍是为在NumPy级别工作的赋值生成一个兼容的对象。|1068962|如果输入中有太多的NaNs，以类似的方式工作可能会更高效/更快，但是使用s.notna()代替，例如：|1068963|import+pandas+as+pd


result+=+pd.Series([[0,+0,+0]]+*+len(s))
result[s.notna()]+=+s[s.notna()]|1068964|让我们尝试做一些基准测试，其中：|1068965|replace_nan_isna()来自于上面|unordered-list-item|1068966|import+pandas+as+pd


def+replace_nan_isna(s,+value,+inplace=False):
++++if+not+inplace:
++++++++s+=+s.copy()
++++ii+=+s.isna()
++++nn+=+ii.sum()
++++s[ii]+=+pd.Series([value]+*+nn).to_numpy()
++++return+s|1068967|replace_nan_notna()也来自于上面|1068968|import+pandas+as+pd


def+replace_nan_notna(s,+value,+inplace=False):
++++if+inplace:
++++++++raise+ValueError("In-place+not+supported!")
++++result+=+pd.Series([value]+*+len(s))
++++result[s.notna()]+=+s[s.notna()]
++++return+result|1068969|replace_nan_reindex()来自@ShubhamSharma的答复|1068970|def+replace_nan_reindex(s,+value,+inplace=False):
++++if+not+inplace:
++++++++s+=+s.copy()
++++s.dropna().reindex(s.index,+fill_value=value)
++++return+s|1068971|replace_nan_fillna()来自@jezrael的回答|1068972|import+pandas+as+pd


def+replace_nan_fillna(s,+value,+inplace=False):
++++if+not+inplace:
++++++++s+=+s.copy()
++++s.fillna(pd.Series([value]+*+len(s),+index=s.index))
++++return+s|1068973|使用以下代码：|1068974|import+numpy+as+np
import+pandas+as+pd


def+gen_data(n=5,+k=2,+p=0.7,+obj=(1,+2,+3)):
++++return+pd.Series(([obj]+*+int(p+*+n)+%2B+[np.nan]+*+(n+-+int(p+*+n)))+*+k)


funcs+=+replace_nan_isna,+replace_nan_notna,+replace_nan_reindex,+replace_nan_fillna

#+:+inspect+results
s+=+gen_data(5,+1)
for+func+in+funcs:
++++print(f'{func.__name__:>20s}++{func(s,+value)}')
print()

#+:+generate+benchmarks
s+=+gen_data(100,+1000)
value+=+(0,+0,+0)
base+=+funcs[0](s,+value)
for+func+in+funcs:
++++print(f'{func.__name__:>20s}++{(func(s,+value)+==+base).all()!s:>5}',+end='++')
++++%25timeit+func(s,+value)
#+++++replace_nan_isna+++True++100+loops,+best+of+5:+16.5+ms+per+loop
#++++replace_nan_notna+++True++10+loops,+best+of+5:+46.5+ms+per+loop
#++replace_nan_reindex+++True++100+loops,+best+of+5:+9.74+ms+per+loop
#+++replace_nan_fillna+++True++10+loops,+best+of+5:+36.4+ms+per+loop|1068975|表示reindex()可能是最快的方法。|1068976|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.fillna.html|1|https://stackoverflow.com/a/72257434/5218354|2|https://stackoverflow.com/a/72257416/5218354^0|9|4|0|2|0|0|37|B|0|0|T|B|16|E|0|0|0|0|0|10|9|0|0|0|0|I|0|0|0|J|0|0|0|L|N|H|1|0|0|0|K|M|B|2|0|0|0|0|2|9|0^^$0|@$1|2|3|4|5|6|7|23|8|@$9|24|A|25|B|C]]|D|@$9|26|A|27|1|28]]|E|$]]|$1|F|3|G|5|H|7|29|8|@$9|2A|A|2B|B|I]]|D|@]|E|$]]|$1|J|3|K|5|6|7|2C|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|2D|8|@$9|2E|A|2F|B|C]|$9|2G|A|2H|B|C]]|D|@]|E|$]]|$1|N|3|O|5|P|7|2I|8|@]|D|@]|E|$Q|R]]|$1|S|3|T|5|6|7|2J|8|@]|D|@]|E|$]]|$1|U|3|V|5|P|7|2K|8|@]|D|@]|E|$Q|R]]|$1|W|3|X|5|6|7|2L|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|2M|8|@$9|2N|A|2O|B|C]]|D|@]|E|$]]|$1|10|3|11|5|P|7|2P|8|@]|D|@]|E|$Q|R]]|$1|12|3|13|5|6|7|2Q|8|@]|D|@]|E|$]]|$1|14|3|15|5|16|7|2R|8|@$9|2S|A|2T|B|C]]|D|@]|E|$]]|$1|17|3|18|5|P|7|2U|8|@]|D|@]|E|$Q|R]]|$1|19|3|1A|5|16|7|2V|8|@$9|2W|A|2X|B|C]]|D|@]|E|$]]|$1|1B|3|1C|5|P|7|2Y|8|@]|D|@]|E|$Q|R]]|$1|1D|3|1E|5|16|7|2Z|8|@$9|30|A|31|B|C]]|D|@$9|32|A|33|1|34]]|E|$]]|$1|1F|3|1G|5|P|7|35|8|@]|D|@]|E|$Q|R]]|$1|1H|3|1I|5|16|7|36|8|@$9|37|A|38|B|C]]|D|@$9|39|A|3A|1|3B]]|E|$]]|$1|1J|3|1K|5|P|7|3C|8|@]|D|@]|E|$Q|R]]|$1|1L|3|1M|5|6|7|3D|8|@]|D|@]|E|$]]|$1|1N|3|1O|5|P|7|3E|8|@]|D|@]|E|$Q|R]]|$1|1P|3|1Q|5|6|7|3F|8|@$9|3G|A|3H|B|C]]|D|@]|E|$]]|$1|1R|3|-4|5|6|7|3I|8|@]|D|@]|E|$]]]|1S|$1T|$5|1U|1V|1W|E|$1X|1Y]]|1Z|$5|1U|1V|1W|E|$1X|20]]|21|$5|1U|1V|1W|E|$1X|22]]]]

The <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.fillna.html" rel="nofollow noreferrer">documentation</a> indicates that this value cannot be a <code>list</code>.
<blockquote>
Value to use to fill holes (e.g. 0), alternately a
dict/Series/DataFrame of values specifying which value to use for each
index (for a Series) or column (for a DataFrame). Values not in the
dict/Series/DataFrame will not be filled. This value cannot be a list.
</blockquote>
This is probably a limitation of the current implementation, and short of patching the source code you must resort to workarounds (as provided below).
<hr />
However, if you are not planning to work with jagged arrays, what you really want to do is probably replace <code>pd.Series()</code> with <code>pd.DataFrame()</code>, e.g.:
<pre class="lang-py prettyprint-override"><code>import numpy as np
import pandas as pd


s = pd.DataFrame(
 [[1, 2, 3],
 [1, 2, 3],
 [np.nan],
 [1, 2, 3],
 [1, 2, 3],
 [np.nan]],
 dtype=pd.Int64Dtype()) # to mix integers with NaNs


s.fillna(0)
# 0 1 2
# 0 1 2 3
# 1 1 2 3
# 2 0 0 0
# 3 1 2 3
# 4 1 2 3
# 5 0 0 0
</code></pre>
<hr />
If you do need to use jagged array, you could use any of the proposed workaround from other answers, or you could make one of your attempt work, e.g.:
<pre class="lang-py prettyprint-override"><code>ii = s.isna()
nn = ii.sum()
s[ii] = pd.Series([[0, 0, 0]] * nn).to_numpy()
# 0 [1, 2, 3]
# 1 [1, 2, 3]
# 2 [0, 0, 0]
# 3 [1, 2, 3]
# 4 [1, 2, 3]
# 5 [0, 0, 0]
# dtype: object
</code></pre>
which basically uses NumPy masking to fill in the Series. The trick is to generate a compatible object for the assignment that works at the NumPy level.
If there are too many NaNs in the input, it is probably more efficient / faster to work in a similar way but with <code>s.notna()</code> instead, e.g.:
<pre class="lang-py prettyprint-override"><code>import pandas as pd


result = pd.Series([[0, 0, 0]] * len(s))
result[s.notna()] = s[s.notna()]
</code></pre>
<hr />
Let's try to do some benchmarking, where:
<ul>
<li><code>replace_nan_isna()</code> is from above</li>
</ul>
<pre class="lang-py prettyprint-override"><code>import pandas as pd


def replace_nan_isna(s, value, inplace=False):
 if not inplace:
 s = s.copy()
 ii = s.isna()
 nn = ii.sum()
 s[ii] = pd.Series([value] * nn).to_numpy()
 return s
</code></pre>
<ul>
<li><code>replace_nan_notna()</code> is also from above</li>
</ul>
<pre><code>import pandas as pd


def replace_nan_notna(s, value, inplace=False):
 if inplace:
 raise ValueError(&quot;In-place not supported!&quot;)
 result = pd.Series([value] * len(s))
 result[s.notna()] = s[s.notna()]
 return result
</code></pre>
<ul>
<li><code>replace_nan_reindex()</code> is from <a href="https://stackoverflow.com/a/72257434/5218354">@ShubhamSharma's answer</a></li>
</ul>
<pre class="lang-py prettyprint-override"><code>def replace_nan_reindex(s, value, inplace=False):
 if not inplace:
 s = s.copy()
 s.dropna().reindex(s.index, fill_value=value)
 return s
</code></pre>
<ul>
<li><code>replace_nan_fillna()</code> is from <a href="https://stackoverflow.com/a/72257416/5218354">@jezrael's answer</a></li>
</ul>
<pre class="lang-py prettyprint-override"><code>import pandas as pd


def replace_nan_fillna(s, value, inplace=False):
 if not inplace:
 s = s.copy()
 s.fillna(pd.Series([value] * len(s), index=s.index))
 return s
</code></pre>
with the following code:
<pre class="lang-py prettyprint-override"><code>import numpy as np
import pandas as pd


def gen_data(n=5, k=2, p=0.7, obj=(1, 2, 3)):
 return pd.Series(([obj] * int(p * n) + [np.nan] * (n - int(p * n))) * k)


funcs = replace_nan_isna, replace_nan_notna, replace_nan_reindex, replace_nan_fillna

# : inspect results
s = gen_data(5, 1)
for func in funcs:
 print(f'{func.__name__:&gt;20s} {func(s, value)}')
print()

# : generate benchmarks
s = gen_data(100, 1000)
value = (0, 0, 0)
base = funcs[0](s, value)
for func in funcs:
 print(f'{func.__name__:&gt;20s} {(func(s, value) == base).all()!s:&gt;5}', end=' ')
 %timeit func(s, value)
# replace_nan_isna True 100 loops, best of 5: 16.5 ms per loop
# replace_nan_notna True 10 loops, best of 5: 46.5 ms per loop
# replace_nan_reindex True 100 loops, best of 5: 9.74 ms per loop
# replace_nan_fillna True 10 loops, best of 5: 36.4 ms per loop
</code></pre>
indicating that <code>reindex()</code> may be the fastest approach.

I have a series like this
<pre class="lang-py prettyprint-override"><code>s = pd.Series([[1,2,3],[1,2,3],np.nan,[1,2,3],[1,2,3],np.nan])
</code></pre>
and I simply want the <code>NaN</code> to be replaced by <code>[0,0,0]</code>.
I have tried
<pre class="lang-py prettyprint-override"><code>s.fillna([0,0,0]) # TypeError: &quot;value&quot; parameter must be a scalar or dict, but you passed a &quot;list&quot;

s[s.isna()] = [[0,0,0],[0,0,0]] # just replaces the NaN with a single &quot;0&quot;. WHY?!

s.fillna(&quot;NAN&quot;).replace({&quot;NAN&quot;:[0,0,0]}) # ValueError: NumPy boolean array indexing assignment cannot 
 #assign 3 input values to the 2 output values where the mask is true


s.fillna(&quot;NAN&quot;).replace({&quot;NAN&quot;:[[0,0,0],[0,0,0]]}) # TypeError: NumPy boolean array indexing assignment
 # requires a 0 or 1-dimensional input, input has 2 dimensions
</code></pre>
I really can't understand, why the two first approaches won't work (maybe I get the first, but the second I cant wrap my head around).
Thanks to <a href="https://stackoverflow.com/questions/31567218/replace-nan-with-empty-list-in-a-pandas-dataframe">this</a> SO-question and answer, we can do it by
<pre class="lang-py prettyprint-override"><code>is_na = s.isna()
s.loc[is_na] = s.loc[is_na].apply(lambda x: [0,0,0])
</code></pre>
but since <code>apply</code> often is rather slow I cannot understand, why we can't use <code>replace</code> or the slicing as above

Why cant Pandas replace nan with an array of 0s using masks/replace?

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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我有这样的系列s = pd.Series([[1,2,3],[1,2,3],np.nan,[1,2,3],[1,2,3],np.nan])我只想让NaN被[0,0,0]取代。我试过了s.fillna([0,0,0]) # TypeError: "value" parameter must be a scalar or ...

问为什么Pandas不能用一个使用掩码/替换的0数组代替nan呢？
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问为什么Pandas不能用一个使用掩码/替换的0数组代替nan呢？
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