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Python Pandas从一列字符串的数据选择中过滤掉Nan

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问题：Python Pandas从一列字符串的数据选择中过滤掉Nan
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问题：Python Pandas从一列字符串的数据选择中过滤掉Nan

如果不使用groupby，我将如何过滤掉没有的数据NaN？

假设我有一个矩阵，客户可以在其中填写“ N / A”，“ n / a”或其任何变体，而其他人则将其留空：

import pandas as pd
import numpy as np


df = pd.DataFrame({'movie': ['thg', 'thg', 'mol', 'mol', 'lob', 'lob'],
                  'rating': [3., 4., 5., np.nan, np.nan, np.nan],
                  'name': ['John', np.nan, 'N/A', 'Graham', np.nan, np.nan]})

nbs = df['name'].str.extract('^(N/A|NA|na|n/a)')
nms=df[(df['name'] != nbs) ]

输出：

>>> nms
  movie    name  rating
0   thg    John       3
1   thg     NaN       4
3   mol  Graham     NaN
4   lob     NaN     NaN
5   lob     NaN     NaN

我将如何过滤NaN值，以便可以像这样获得结果：

  movie    name  rating
0   thg    John       3
3   mol  Graham     NaN

我猜我需要类似的东西，~np.isnan但tilda不适用于字符串。

点击查看英文原文

Without using groupby how would I filter out data without NaN?

Let say I have a matrix where customers will fill in ‘N/A’,’n/a’ or any of its variations and others leave it blank:

import pandas as pd
import numpy as np


df = pd.DataFrame({'movie': ['thg', 'thg', 'mol', 'mol', 'lob', 'lob'],
                  'rating': [3., 4., 5., np.nan, np.nan, np.nan],
                  'name': ['John', np.nan, 'N/A', 'Graham', np.nan, np.nan]})

nbs = df['name'].str.extract('^(N/A|NA|na|n/a)')
nms=df[(df['name'] != nbs) ]

output:

>>> nms
  movie    name  rating
0   thg    John       3
1   thg     NaN       4
3   mol  Graham     NaN
4   lob     NaN     NaN
5   lob     NaN     NaN

How would I filter out NaN values so I can get results to work with like this:

  movie    name  rating
0   thg    John       3
3   mol  Graham     NaN

I am guessing I need something like ~np.isnan but the tilda does not work with strings.

回答 0

只需将它们放下：

nms.dropna(thresh=2)

这将删除所有至少有两个non-的行NaN。

然后，您可以删除名称为NaN：

In [87]:

nms
Out[87]:
  movie    name  rating
0   thg    John       3
1   thg     NaN       4
3   mol  Graham     NaN
4   lob     NaN     NaN
5   lob     NaN     NaN

[5 rows x 3 columns]
In [89]:

nms = nms.dropna(thresh=2)
In [90]:

nms[nms.name.notnull()]
Out[90]:
  movie    name  rating
0   thg    John       3
3   mol  Graham     NaN

[2 rows x 3 columns]

编辑

实际查看您最初想要的是什么，而无需dropna调用即可：

nms[nms.name.notnull()]

更新

3年后的这个问题，有一个错误，首先thresharg至少查找n非NaN值，因此实际上输出应为：

In [4]:
nms.dropna(thresh=2)

Out[4]:
  movie    name  rating
0   thg    John     3.0
1   thg     NaN     4.0
3   mol  Graham     NaN

我可能是3年前弄错了，或者我运行的熊猫版本存在错误，两种情况都是可能的。

点击查看英文原文

Just drop them:

nms.dropna(thresh=2)

this will drop all rows where there are at least two non-NaN.

Then you could then drop where name is NaN:

In [87]:

nms
Out[87]:
  movie    name  rating
0   thg    John       3
1   thg     NaN       4
3   mol  Graham     NaN
4   lob     NaN     NaN
5   lob     NaN     NaN

[5 rows x 3 columns]
In [89]:

nms = nms.dropna(thresh=2)
In [90]:

nms[nms.name.notnull()]
Out[90]:
  movie    name  rating
0   thg    John       3
3   mol  Graham     NaN

[2 rows x 3 columns]

EDIT

Actually looking at what you originally want you can do just this without the dropna call:

nms[nms.name.notnull()]

UPDATE

Looking at this question 3 years later, there is a mistake, firstly thresh arg looks for at least n non-NaN values so in fact the output should be:

In [4]:
nms.dropna(thresh=2)

Out[4]:
  movie    name  rating
0   thg    John     3.0
1   thg     NaN     4.0
3   mol  Graham     NaN

It’s possible that I was either mistaken 3 years ago or that the version of pandas I was running had a bug, both scenarios are entirely possible.

回答 1

所有解决方案中最简单的：

filtered_df = df[df['name'].notnull()]

因此，它仅过滤掉“名称”列中没有NaN值的行。

对于多列：

filtered_df = df[df[['name', 'country', 'region']].notnull().all(1)]
点击查看英文原文

Simplest of all solutions:

filtered_df = df[df['name'].notnull()]

Thus, it filters out only rows that doesn’t have NaN values in ‘name’ column.

For multiple columns:

filtered_df = df[df[['name', 'country', 'region']].notnull().all(1)]

回答 2

df = pd.DataFrame({'movie': ['thg', 'thg', 'mol', 'mol', 'lob', 'lob'],'rating': [3., 4., 5., np.nan, np.nan, np.nan],'name': ['John','James', np.nan, np.nan, np.nan,np.nan]})

for col in df.columns:
    df = df[~pd.isnull(df[col])]
点击查看英文原文 
df = pd.DataFrame({'movie': ['thg', 'thg', 'mol', 'mol', 'lob', 'lob'],'rating': [3., 4., 5., np.nan, np.nan, np.nan],'name': ['John','James', np.nan, np.nan, np.nan,np.nan]})

for col in df.columns:
    df = df[~pd.isnull(df[col])]

回答 3

df.dropna(subset=['columnName1', 'columnName2'])
点击查看英文原文 
df.dropna(subset=['columnName1', 'columnName2'])