Python 使用 str.contains 忽略 NaN
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Ignoring NaNs with str.contains
提问by Emre
I want to find rows that contain a string, like so:
我想查找包含字符串的行，如下所示：
DF[DF.col.str.contains("foo")]
However, this fails because some elements are NaN:
但是，这失败了，因为有些元素是 NaN：
ValueError: cannot index with vector containing NA / NaN values
ValueError: 无法索引包含 NA / NaN 值的向量
So I resort to the obfuscated
所以我求助于混淆
DF[DF.col.notnull()][DF.col.dropna().str.contains("foo")]
Is there a better way?
有没有更好的办法？
采纳答案by Andy Hayden
There's a flag for that:
有一个标志：
In [11]: df = pd.DataFrame([["foo1"], ["foo2"], ["bar"], [np.nan]], columns=['a'])
In [12]: df.a.str.contains("foo")
Out[12]:
0 True
1 True
2 False
3 NaN
Name: a, dtype: object
In [13]: df.a.str.contains("foo", na=False)
Out[13]:
0 True
1 True
2 False
3 False
Name: a, dtype: bool
See the str.replacedocs:
查看str.replace文档：
na : default NaN, fill value for missing values.
na : 默认 NaN，缺失值的填充值。
So you can do the following:
因此，您可以执行以下操作：
In [21]: df.loc[df.a.str.contains("foo", na=False)]
Out[21]:
a
0 foo1
1 foo2
回答by Faheem Alvi
import folium
import pandas
data= pandas.read_csv("maps.txt")
lat = list(data["latitude"])
lon = list(data["longitude"])
map= folium.Map(location=[31.5204, 74.3587], zoom_start=6, tiles="Mapbox Bright")
fg = folium.FeatureGroup(name="My Map")
for lt, ln in zip(lat, lon):
c1 = fg.add_child(folium.Marker(location=[lt, ln], popup="Hi i am a Country",icon=folium.Icon(color='green')))
child = fg.add_child(folium.Marker(location=[31.5204, 74.5387], popup="Welcome to Lahore", icon= folium.Icon(color='green')))
map.add_child(fg)
map.save("Lahore.html")
Traceback (most recent call last):
File "C:\Users\Ryan\AppData\Local\Programs\Python\Python36-32\check2.py", line 14, in <module>
c1 = fg.add_child(folium.Marker(location=[lt, ln], popup="Hi i am a Country",icon=folium.Icon(color='green')))
File "C:\Users\Ryan\AppData\Local\Programs\Python\Python36-32\lib\site-packages\folium\map.py", line 647, in __init__
self.location = _validate_coordinates(location)
File "C:\Users\Ryan\AppData\Local\Programs\Python\Python36-32\lib\site-packages\folium\utilities.py", line 48, in _validate_coordinates
'got:\n{!r}'.format(coordinates))
ValueError: Location values cannot contain NaNs, got:
[nan, nan]
回答by Harry_pb
In addition to the above answers, I would say for columns having no single word name, you may use:-
除了上述答案，我想说对于没有单个单词名称的列，您可以使用：-
df[df['Product ID'].str.contains("foo") == True]
Hope this helps.
希望这可以帮助。
回答by Nate Taylor
I'm not 100% on why (actually came here to search for the answer), but this also works, and doesn't require replacing all nan values.
我不是 100% 为什么（实际上是来这里寻找答案），但这也有效，并且不需要替换所有 nan 值。
import pandas as pd
import numpy as np
df = pd.DataFrame([["foo1"], ["foo2"], ["bar"], [np.nan]], columns=['a'])
newdf = df.loc[df['a'].str.contains('foo') == True]
Works with or without .loc.
有或没有.loc.
I have no idea why this works, as I understand it when you're indexing with brackets pandas evaluates whatever's inside the bracket as either Trueor False. I can't tell why making the phrase inside the brackets 'extra boolean' has any effect at all.
我不知道为什么会这样，正如我所理解的，当您使用括号进行索引时，pandas 会将括号内的任何内容评估为Trueor 或False。我不知道为什么将括号内的短语设为“额外布尔值”会产生任何影响。
回答by Aliakbar Hosseinzadeh
You can also patern :
你也可以模式：
DF[DF.col.str.contains(pat = '(foo)', regex = True) ]
