pandas 如何在熊猫的字符串列中按 NaN 过滤？
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How to filter by NaN in string column in pandas?
提问by Richard
I'm using pandas 0.18. I have loaded a dataframe from CSV using pd.read_csv(), and it looks as thought the empty cells in CSV have loaded as NaNin the dataframe.
我正在使用Pandas 0.18。我已经使用 CSV 从 CSV 加载了一个数据框pd.read_csv()，看起来 CSV 中的空单元格已经像NaN在数据框中一样加载了。
Now I want to find the number of rows with an empty value in a particular column, but I'm struggling.
现在我想在特定列中找到具有空值的行数，但我很挣扎。
This is my dataframe:
这是我的数据框：
ods id provider
0 A86016 NaN emis
1 L81042 463061 NaN
2 C84013 NaN tpp
3 G82228 462941 emis
4 C81083 NaN tpp
This is what I get from a df.describe():
这是我从一个df.describe()：
ods id provider
count 9897 7186 9022
unique 8066 192 4
top N83028 463090 emis
freq 7 169 4860
I want to get all the rows where providerwas empty in the CSV. This is what I've tried:
我想获取providerCSV 中所有为空的行。这是我尝试过的：
>>> print len(df[df.provider == 'NaN'])
0
>>> print len(df[df.provider == np.nan])
0
I can see that there are some NaNvalues in there (e.g. row 1) so what gives?
我可以看到那里有一些NaN值（例如第 1 行）那么是什么给出的？
Also, why does pandas convert empty values in string columns like providerto NaN- wouldn't it make more sense to convert them to an empty string?
此外，为什么Pandas转换为空值字符串列样provider来NaN-那岂不是更有意义，将它们转换为空字符串？
回答by jezrael
Use isnullfor comparing NaN:
使用isnull比较NaN：
df = pd.DataFrame({'ods': {0: 'A86016', 1: 'L81042', 2: 'C84013', 3: 'G82228', 4: 'C81083'},
'id': {0: np.nan, 1: 463061.0, 2: np.nan, 3: 462941.0, 4: np.nan},
'provider': {0: 'emis', 1: np.nan, 2: 'tpp', 3: 'emis', 4: 'tpp'}})
print df
id ods provider
0 NaN A86016 emis
1 463061.0 L81042 NaN
2 NaN C84013 tpp
3 462941.0 G82228 emis
4 NaN C81083 tpp
print (df[df.provider.isnull()])
ods id provider
1 L81042 463061.0 NaN
print len(df[df.provider.isnull()])
1
If you need convert NaNto `` use fillna:
如果您需要转换NaN为 `` 使用fillna：
df.provider.fillna('', inplace=True)
print df
id ods provider
0 NaN A86016 emis
1 463061.0 L81042
2 NaN C84013 tpp
3 462941.0 G82228 emis
4 NaN C81083 tpp
Docs:
文档：
Warning
One has to be mindful that in python (and numpy), the nan's don't compare equal, but None's do. Note that Pandas/numpy uses the fact that np.nan != np.nan, and treats None like np.nan.
警告
必须注意，在 python（和 numpy）中，nan 不相等，但 None 不相等。请注意，Pandas/numpy 使用 np.nan != np.nan 的事实，并将 None 视为 np.nan。
In [11]: None == None
Out[11]: True
In [12]: np.nan == np.nan
Out[12]: False
So as compared to above, a scalar equality comparison versus a None/np.nan doesn't provide useful information.
因此，与上面相比，标量相等比较与 None/np.nan 没有提供有用的信息。
In [13]: df2['one'] == np.nan
Out[13]:
a False
b False
c False
d False
e False
f False
g False
h False
Name: one, dtype: bool
But if nanis string:
但如果nan是字符串：
df = pd.DataFrame({'ods': {0: 'A86016', 1: 'L81042', 2: 'C84013', 3: 'G82228', 4: 'C81083'},
'id': {0: np.nan, 1: 463061.0, 2: np.nan, 3: 462941.0, 4: np.nan},
'provider': {0: 'emis', 1: 'nan', 2: 'tpp', 3: 'emis', 4: 'tpp'}})
print df
ods id provider
0 A86016 NaN emis
1 L81042 463061.0 nan
2 C84013 NaN tpp
3 G82228 462941.0 emis
4 C81083 NaN tpp
print (df[df.provider == 'nan'])
ods id provider
1 L81042 463061.0 nan
do you know why pandas imports empty strings as NaN rather than empty strings?
你知道为什么 Pandas 将空字符串作为 NaN 而不是空字符串导入吗？
See docs(bold by me):
查看文档（我加粗）：
na_values : str, list-likeor dict, default None
Additional strings to recognize as NA/NaN. If dict passed, specific per-column NA values. By default the following values are interpreted as NaN: '-1.#IND', '1.#QNAN', '1.#IND', '-1.#QNAN', '#N/A N/A', '#N/A', 'N/A', 'NA', '#NA', 'NULL', 'NaN', '-NaN', 'nan', '-nan', ''.
na_values : str, list-like或dict, 默认无
要识别为 NA/NaN 的其他字符串。如果 dict 通过，特定的每列 NA 值。默认情况下，以下值被解释为 NaN：'-1.#IND'、'1.#QNAN'、'1.#IND'、'-1.#QNAN'、'#N/AN/A'、' #N/A', 'N/A', 'NA', '#NA', 'NULL', 'NaN', '-NaN', 'nan', '-nan', ''。
回答by ysearka
You can first store the na values, and then drop all the rest:
您可以先存储 na 值，然后删除所有其余值：
without_na = df['provider'].dropna()
df[~df.index.isin(without_na.index)]
