Develop Reference

Python numpy array filter

I got the following numpy array named 'data'. It consists of 15118 rows and 2 columns. The first column mostly consist of 0.01 steps, but sometimes there is a step in between (shown in red) which I would like to remove/filter out.
I achieved this with the following code:
# Create array [0, 0.01 .... 140], rounded 2 decimals to prevent floating point error
b = np.round(np.arange(0,140.01,0.01),2)
# New empty data array
new_data = np.empty(shape=[0, 2])
# Loop over values to remove/filter out data
for x in b:
Index = np.where(x == data[:,0])[0][0]
new_data = np.vstack([new_data,data[Index]])
I feel like this code is far from optimal and I was wondering if anyone knows a faster/better way of achieving this?

Here's a solution using pandas for resampling, you can probably achieve the same result in pure numpy but there are a number of floating point and rounding error pitfalls you are going to face, maybe it's better to let a trusted library do the work for you.
Let's say arr is your data array and assume your index to be in fractions of seconds. You can convert your array to a dataframe with a timedelta index:
df = pd.DataFrame(arr[:,1], index=arr[:,0])
df.index = pd.to_timedelta(df.index, unit="s")
Than resampling it's pretty easy, 10ms is the frequency you want, first() should give you the expected result dropping everything but the records at 10ms ticks, but feel free to experiment with other functions
df = df.resample("10ms").first()
Eventually you could get back to your array with something like:
np.vstack([pd.to_numeric(df.index, downcast="float").values / 1e9,
df.values.squeeze()]).T

Find big differences in numpy array

I have a csv file that contains data from two led measurements. There are some mistakes in the file that gives huge sparks in the graph. I want to locate this places where this happens.
I have this code that makes two arrays that I plot.
x625 = np.array(df['LED Group 625'].dropna(axis=0, how='all'))
x940 = np.array(df['LED Group 940'].dropna(axis=0, how='all'))

I will provide an answer with some artificial data since you have not posted any data yet.
So after you convert the pandas columns into a numpy array, you can do something like this:
import numpy as np
# some random data. 100 lines and 1 column
x625 = np.random.rand(100,1)
# Assume that the maximum value in `x625` is a spark.
spark = x625.max()
# Find where these spark are in the `x625`
np.where(x625==spark)
#(array([64]), array([0]))
The above means that a value equal to spark is located on the 64th line of the 0th column.
Similarly, you can use np.where(x625 > any_number_here)
If instead of the location you need to create a boolean mask use this:
boolean_mask = (x625==spark)
# verify
np.where(boolean_mask)
# (array([64]), array([0]))
EDIT 1
You can use numpy.diff() to get the element wise differences of all elements into the list (variable).
diffs = np.diff(x625.ravel())
This will have in index 0 the results of element1-element0.
If the vaules in diffs are big in a specific index positio, then a spark occured in that position.

Getting the index of a float in a column using pandas

I have a dataset that I am pulling using pandas. It looks like this:
import pandas as pd
dataset=pd.read_csv('D:\\filename.csv', header=None, usecols=3,4,10,16,22,28])
time=dataset.iloc[:,0]
Now, the 'time' dataset has a value of 0.00017 somewhere down the column and I want to find the index number of that location. How can I get that?

Assuming you're dealing with floats, you can't use an equality comparison here (because of floating point inaccuracies creeping in).
Use np.isclose + np.argmax:
idx = np.isclose(df['time'], 0.00017).argmax()
If there's a possibility this value may not exist:
m = np.isclose(df['time'], 0.00017)
if m.sum() > 0:
idx = m.argmax()
Otherwise, set idx to whatever (None, -1, etc).

Trying to divide a dataframe column by a float yields NaN

Background
I deal with a csv datasheet that prints out columns of numbers. I am working on a program that will take the first column, ask a user for a time in float (ie. 45 and a half hours = 45.5) and then subtract that number from the first column. I have been successful in that regard. Now, I need to find the row index of the "zero" time point. I use min to find that index and then call that off of the following column A1. I need to find the reading at Time 0 to then normalize A1 to so that on a graph, at the 0 time point the reading is 1 in column A1 (and eventually all subsequent columns but baby steps for me)
time_zero = float(input("Which time would you like to be set to 0?"))
df['A1']= df['A1']-time_zero
This works fine so far to set the zero time.
zero_location_series = df[df['A1'] == df['A1'].min()]
r1 = zero_location_series[' A1.1']
df[' A1.1'] = df[' A1.1']/r1
Here's where I run into trouble. The first line will correctly identify a series that I can pull off of for all my other columns. Next r1 correctly identifies the proper A1.1 value and this value is a float when I use type(r1).
However when I divide df[' A1.1']/r1 it yields only one correct value and that value is where r1/r1 = 1. All other values come out NaN.
My Questions:
How to divide a column by a float I guess? Why am I getting NaN?
Is there a faster way to do this as I need to do this for 16 columns.(ie 'A2/r2' 'a3/r3' etc.)
Do I need to do inplace = True anywhere to make the operations stick prior to resaving the data? or is that only for adding/deleting rows?
Example
Dataframe that looks like this
!http://i.imgur.com/ObUzY7p.png
zero time sets properly (image not shown)
after dividing the column
!http://i.imgur.com/TpLUiyE.png

This should work:
df['A1.1']=df['A1.1']/df['A1.1'].min()
I think the reason df[' A1.1'] = df[' A1.1']/r1 did not work was because r1 is a series. Try r1? instead of type(r1) and pandas will tell you that r1 is a series, not an individual float number.
To do it in one attempt, you have to iterate over each column, like this:
for c in df:
df[c] = df[c]/df[c].min()

If you want to divide every value in the column by r1 it's best to apply, for example:
import pandas as pd
df = pd.DataFrame([1,2,3,4,5])
# apply an anonymous function to the first column ([0]), divide every value
# in the column by 3
df = df[0].apply(lambda x: x/3.0, 0)
print(df)
So you'd probably want something like this:
df = df["A1.1"].apply(lambda x: x/r1, 0)
This really only answers part 2 of you question. Apply is probably your best bet for running a function on multiple rows and columns quickly. As for why you're getting nans when dividing by a float, is it possible the values in your columns are anything other than floats or integers?

Pandas python barplot by subgroup

Ok so I have a dataframe object that's indexed as follows:
index, rev, metric1 (more metrics.....)
exp1, 92365, 0.018987
exp2, 92365, -0.070901
exp3, 92365, 0.150140
exp1, 87654, 0.003008
exp2, 87654, -0.065196
exp3, 87654, -0.174096
For each of these metrics I want to create individual stacked barplots comparing them based on their rev.
here's what I've tried:
df = df[['rev', 'metric1']]
df = df.groupby("rev")
df.plot(kind = 'bar')
This results in 2 individual bar graphs of the metric. Ideally I would have these two merged and stacked (right now stacked=true does nothing). Any help would be much appreciated.
This would give me my ideal result, however I don't think reorganizing to fit this is the best way to achieve my goal as I have many metrics and many revisions.
index, metric1(rev87654), metric1(rev92365)
exp1, 0.018987, 0.003008
exp2, -0.070901, -0.065196
exp3, 0.150140, -0.174096
This is my goal. (made by hand)
http://i.stack.imgur.com/5GRqB.png

following from this matplotlib gallery example:
http://matplotlib.org/examples/api/barchart_demo.html
there they get multiple to plot by calling bar once for each set.
You could access these values in pandas with indexing operations as follows:
fig, ax = subplots(figsize=(16.2,10),dpi=300)
Y = Tire2[Tire2.SL==Tire2.SL.unique()[0]].SA.values[0:13]
X = linspace(0,size(Y),size(Y))
ax.bar(X,Y,width=.4)
Y = Tire2[Tire2.SL==Tire2.SL.unique()[2]].SA.values[0:13]
X = linspace(0,size(Y),size(Y))+.5
ax.bar(X,Y,width=.4,color='r')
working from the inside out:
get all of the unique values of 'SL' in one of the cols (rev in your case)
Get a Boolean vector of all rows where 'SL' equals the first (or nth) unique value
Index Tire by that Boolean vector (this will pull out only those rows where the vector is True
access the values of SA or a metric in yourcase. (took only the `[0:13]' values because i was testing this on a huge data set)
bar plot those values
if your experiments are consistently in order in the frame(as shown), that's that. Otherwise you might need to run a little sorting to get your Y values in the right order. .sort(column name) should take care of that. In my code, i'd slip it in between ...[0]] and.SA...
In general, this kind of operation can really help you out in wrangling big frames. .between is useful. And you can always add, multiply etc. the Boolean vectors to construct more complex logic.

I'm not sure how to get the plot you want automatically without doing exactly the reorganization you specify at the end. The answer by user3823992 gives you more detailed control of the plots, but if you want them more automatic here is some temporary reorganization that should work using the indexing similarly but also concatenating back into a DataFrame that will do the plot for you.
import numpy as np
import pandas as pd
exp = ['exp1','exp2','exp3']*2
rev = [1,1,1,2,2,2]
met1 = np.linspace(-0.5,1,6)
met2 = np.linspace(1.0,5.0,6)
met3 = np.linspace(-1,1,6)
df = pd.DataFrame({'rev':rev, 'met1':met1, 'met2':met2, 'met3':met3}, index=exp)
for met in df.columns:
if met != 'rev':
merged = df[df['rev'] == df.rev.unique()[0]][met]
merged.name = merged.name+'rev'+str(df.rev.unique()[0])
for rev in df.rev.unique()[1:]:
tmp = df[df['rev'] == rev][met]
tmp.name = tmp.name+'rev'+str(rev)
merged = pd.concat([merged, tmp], axis=1)
merged.plot(kind='bar')
This should give you three plots, one for each of my fake metrics.
EDIT : Or something like this might do also
df['exp'] = df.index
pt = pd.pivot_table(df, values='met1', rows=['exp'], cols=['rev'])
pt.plot(kind='bar')