blokeley - 3 months ago 26

Python Question

I have a

`DataFrame`

`DataFrame`

How do I calculate the work done (which is "the area under the curve")?

I looked at

`numpy.trapz`

`import numpy as np`

import pandas as pd

forces = pd.read_csv(...)

work_done = {}

for col in forces.columns:

work_done[col] = np.trapz(forces.loc[col], forces.index))

I was hoping to create a new

`DataFrame`

`dict`

`DataFrame.apply()`

In short:

- Can I avoid the looping?
- Can I create a of work done directly?
`DataFrame`

Thanks in advance for any help.

Answer

You could vectorize this by passing the whole `DataFrame`

to `np.trapz`

and specifying the `axis=`

argument, e.g.:

```
import numpy as np
import pandas as pd
# some random input data
gen = np.random.RandomState(0)
x = gen.randn(100, 10)
names = [chr(97 + i) for i in range(10)]
forces = pd.DataFrame(x, columns=names)
# vectorized version
wrk = np.trapz(forces, x=forces.index, axis=0)
work_done = pd.DataFrame(wrk[None, :], columns=forces.columns)
# non-vectorized version for comparison
work_done2 = {}
for col in forces.columns:
work_done2.update({col:np.trapz(forces.loc[:, col], forces.index)})
```

These give the following output:

```
from pprint import pprint
pprint(work_done.T)
# 0
# a -24.331560
# b -10.347663
# c 4.662212
# d -12.536040
# e -10.276861
# f 3.406740
# g -3.712674
# h -9.508454
# i -1.044931
# j 15.165782
pprint(work_done2)
# {'a': -24.331559643023006,
# 'b': -10.347663159421426,
# 'c': 4.6622123535050459,
# 'd': -12.536039649161403,
# 'e': -10.276861220217308,
# 'f': 3.4067399176289994,
# 'g': -3.7126739591045541,
# 'h': -9.5084536839888187,
# 'i': -1.0449311137294459,
# 'j': 15.165781517623724}
```

There are a couple of other problems with your original example. `col`

is a column name rather than a row index, so it needs to index the second dimension of your dataframe (i.e. `.loc[:, col]`

rather than `.loc[col]`

). Also, you have an extra trailing parenthesis on the last line.

You *could* also generate the output `DataFrame`

directly by `.apply`

ing `np.trapz`

to each column, e.g.:

```
work_done = forces.apply(np.trapz, axis=0, args=(forces.index,))
```

However, this isn't really 'proper' vectorization - you are still calling `np.trapz`

separately on each column. You can see this by comparing the speed of the `.apply`

version against calling `np.trapz`

directly:

```
In [1]: %timeit forces.apply(np.trapz, axis=0, args=(forces.index,))
1000 loops, best of 3: 582 µs per loop
In [2]: %timeit np.trapz(forces, x=forces.index, axis=0)
The slowest run took 6.04 times longer than the fastest. This could mean that an
intermediate result is being cached
10000 loops, best of 3: 53.4 µs per loop
```

This isn't an entirely fair comparison, since the second version excludes the extra time taken to construct the `DataFrame`

from the output numpy array, but this should still be smaller than the difference in time taken to perform the actual integration.