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| Boilerpipe Text | In general you should probably have a look at the following answers:
https://stackoverflow.com/a/13434501/604048
https://stackoverflow.com/a/24489602/604048
However, I'd like to address number 3.
You should use
numpy
to check for
NaN
s, not use the equals operator.
np.isnan(...)
.
You shouldn't use a
while
loop. It's very unpythonic.
Your if statement shouldn't have a parenthesis.
The indentation of
while
part should make this script fail.
Remember you're working with rows. You shouldn't get the ambiguous truth value error if your code was implemented the way you showed.
Maybe you wrote something like this:
i =
0
while
(i <
len
(y)):
if
np.isnan(y[i]):
y[i] =
0
i = i +
1
This produces:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
What is happening here is that the if statement gets a list of bools, and doesn't know what to do with it. Is it true when all of them are true? Or, is it true when any of them are true? You could resolve it as follows:
i =
0
while
(i <
len
(y)):
if
np.isnan(y[i]).
any
():
y[i] =
0
i = i +
1
Or more pythonic:
for
i
in
range
(
0
, len(y)):
if np.
isnan
(y[i]).
any
():
y[i] =
0
Also be careful about the zero assignment. I don't know how pandas data frames are implemented, but typically with normal collection types you'd be replacing a row with the number zero (0).
If we weren't working with special pandas data frames, and it was simply lists I'd go for:
y
=
[
row
for
row
in
y if
not
np.isnan(
row
).
any
()]
... if you wanted to delete a row with any
NaN
values.
Or, if you simply want to set any element which is
NaN
to zero:
y = [[
0
if
np.
isnan
(elm)
else
elm
for
elm
in
row]
for
row
in
y]
With test data:
import
numpy
as
np
y = [[
0
,
0
],[
1
,
1
],[
2
,
float
(
'nan'
)],[
3
,
3
]]
print
(y)
x = [row
for
row
in
y
if
not
np.isnan(row).
any
()]
print
(x)
z = [[
0
if
np.isnan(elm)
else
elm
for
elm
in
row]
for
row
in
y]
print
(z)
This produces the following output:
[[0, 0], [1, 1], [2, nan], [3, 3]]
[[0, 0], [1, 1], [3, 3]]
[[0, 0], [1, 1], [2, 0], [3, 3]] |
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# [Removing NaN Values in Python Quantopian](https://quant.stackexchange.com/questions/31094/removing-nan-values-in-python-quantopian)
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I am trying to make a histogram in numpy but numpy.histogram seems to really hate NaN values. I have tried removing NaN values from a list called data in three different ways and Quantopian doesn't let me use any of those three ways: 1.) TypeError: only integer arrays with one element can be converted to an index
```
data = data[~np.isnan(data)]
```
2\.) So then I tried using pandas.dropna() and it threw: ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
```
datapd = pd.Series(data)
hist, bins = np.histogram(datapd.dropna(), density=True, bins = 'auto')
```
3\.) And when that didn't work I tried removing them on my own:
```
i = 0
while(i < len(y)):
if(float('nan') == y[i]):
y[i] = 0
i = i + 1
```
It threw the same error: ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
How do you remove NaN values in quantopian? I tried these methods in a python console and it worked I have no clue as to why it wouldn't work in quantopian.
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## 1 Answer 1
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In general you should probably have a look at the following answers:
- <https://stackoverflow.com/a/13434501/604048>
- <https://stackoverflow.com/a/24489602/604048>
However, I'd like to address number 3.
1. You should use `numpy` to check for `NaN`s, not use the equals operator. `np.isnan(...)`.
2. You shouldn't use a `while` loop. It's very unpythonic.
3. Your if statement shouldn't have a parenthesis.
4. The indentation of `while` part should make this script fail.
5. Remember you're working with rows. You shouldn't get the ambiguous truth value error if your code was implemented the way you showed.
Maybe you wrote something like this:
```
i = 0
while(i < len(y)):
if np.isnan(y[i]):
y[i] = 0
i = i + 1
```
This produces:
> ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
What is happening here is that the if statement gets a list of bools, and doesn't know what to do with it. Is it true when all of them are true? Or, is it true when any of them are true? You could resolve it as follows:
```
i = 0
while(i < len(y)):
if np.isnan(y[i]).any():
y[i] = 0
i = i + 1
```
Or more pythonic:
```
for i in range(0, len(y)):
if np.isnan(y[i]).any():
y[i] = 0
```
Also be careful about the zero assignment. I don't know how pandas data frames are implemented, but typically with normal collection types you'd be replacing a row with the number zero (0).
If we weren't working with special pandas data frames, and it was simply lists I'd go for:
```
y = [row for row in y if not np.isnan(row).any()]
```
... if you wanted to delete a row with any `NaN` values.
Or, if you simply want to set any element which is `NaN` to zero:
```
y = [[0 if np.isnan(elm) else elm for elm in row] for row in y]
```
With test data:
```
import numpy as np
y = [[0,0],[1,1],[2,float('nan')],[3,3]]
print(y)
x = [row for row in y if not np.isnan(row).any()]
print(x)
z = [[0 if np.isnan(elm) else elm for elm in row] for row in y]
print(z)
```
This produces the following output:
```
[[0, 0], [1, 1], [2, nan], [3, 3]]
[[0, 0], [1, 1], [3, 3]]
[[0, 0], [1, 1], [2, 0], [3, 3]]
```
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| Readable Markdown | In general you should probably have a look at the following answers:
- <https://stackoverflow.com/a/13434501/604048>
- <https://stackoverflow.com/a/24489602/604048>
However, I'd like to address number 3.
1. You should use `numpy` to check for `NaN`s, not use the equals operator. `np.isnan(...)`.
2. You shouldn't use a `while` loop. It's very unpythonic.
3. Your if statement shouldn't have a parenthesis.
4. The indentation of `while` part should make this script fail.
5. Remember you're working with rows. You shouldn't get the ambiguous truth value error if your code was implemented the way you showed.
Maybe you wrote something like this:
```
i = 0
while(i < len(y)):
if np.isnan(y[i]):
y[i] = 0
i = i + 1
```
This produces:
> ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
What is happening here is that the if statement gets a list of bools, and doesn't know what to do with it. Is it true when all of them are true? Or, is it true when any of them are true? You could resolve it as follows:
```
i = 0
while(i < len(y)):
if np.isnan(y[i]).any():
y[i] = 0
i = i + 1
```
Or more pythonic:
```
for i in range(0, len(y)):
if np.isnan(y[i]).any():
y[i] = 0
```
Also be careful about the zero assignment. I don't know how pandas data frames are implemented, but typically with normal collection types you'd be replacing a row with the number zero (0).
If we weren't working with special pandas data frames, and it was simply lists I'd go for:
```
y = [row for row in y if not np.isnan(row).any()]
```
... if you wanted to delete a row with any `NaN` values.
Or, if you simply want to set any element which is `NaN` to zero:
```
y = [[0 if np.isnan(elm) else elm for elm in row] for row in y]
```
With test data:
```
import numpy as np
y = [[0,0],[1,1],[2,float('nan')],[3,3]]
print(y)
x = [row for row in y if not np.isnan(row).any()]
print(x)
z = [[0 if np.isnan(elm) else elm for elm in row] for row in y]
print(z)
```
This produces the following output:
```
[[0, 0], [1, 1], [2, nan], [3, 3]]
[[0, 0], [1, 1], [3, 3]]
[[0, 0], [1, 1], [2, 0], [3, 3]]
``` |
| Shard | 18 (laksa) |
| Root Hash | 8045678284012640218 |
| Unparsed URL | com,stackexchange!quant,/questions/31094/removing-nan-values-in-python-quantopian s443 |