import pandas as pd
import numpy as np
data = 'filename.csv'
df = pd.DataFrame(data)
df
one two three four five
a 0.469112 -0.282863 -1.509059 bar True
b 0.932424 1.224234 7.823421 bar False
c -1.135632 1.212112 -0.173215 bar False
d 0.232424 2.342112 0.982342 unbar True
e 0.119209 -1.044236 -0.861849 bar True
f -2.104569 -0.494929 1.071804 bar False
I would like to select a range for a certain column, let's say column two
. I would like to select all values between -0.5 and +0.5. How does one do this?
I expected to use
-0.5 < df["two"] < 0.5
But this (naturally) gives a ValueError:
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
I tried
-0.5 (< df["two"] < 0.5)
But this outputs all True
.
The correct output should be
0 True
1 False
2 False
3 False
4 False
5 True
What is the correct way to find a range of values in a pandas dataframe column?
EDIT: Question
Using .between()
with
df['two'].between(-0.5, 0.5, inclusive=False)
would would be the difference between
-0.5 < df['two'] < 0.5
and inequalities like
-0.5 =< df['two'] < 0.5
?
To select the rows, the syntax is df. loc[start:stop:step] ; where start is the name of the first-row label to take, stop is the name of the last row label to take, and step as the number of indices to advance after each extraction; for example, you can use it to select alternate rows.
Use between
with inclusive=False
for strict inequalities:
df['two'].between(-0.5, 0.5, inclusive=False)
The inclusive
parameter determines if the endpoints are included or not (True
: <=
, False
: <
). This applies to both signs. If you want mixed inequalities, you'll need to code them explicitly:
(df['two'] >= -0.5) & (df['two'] < 0.5)
.between
is a good solution, but if you want finer control use this:
(0.5 <= df['two']) & (df['two'] < 0.5)
The operator &
is different from and
. The other operators are |
for or
, ~
for not
. See this discussion for more info.
Your statement was the same as this:
(0.5 <= df['two']) and (df['two'] < 0.5)
Hence it raised the error.
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