Python Pandas Pyhaystack

I am using a module called pyhaystack to retrieve data (rest API) from a building automation system based on 'tags.' Python will return a dictionary of the data. Im trying to use p

Solution 1:

I think DataFrame would be a good way to handle what you want. Starting with znt you can make all the calculation there :

deviation = znt - 70
deviation = deviation.abs()

# and the cool part is filtering in the df
problem_zones = 
    deviation[deviation['C.Drivers.NiagaraNetwork.Adams_Friendship.points.A-
    Section.AV1.AV1ZN~2dT']>4]

You can play with this and figure out a way to iterate through columns, like :

foreach in df.columns:
    # if in this column, more than 10 occurences of deviation GT 4...if len(df[df[each]>4]) > 10:
        print('This zone have a lot of troubles : ', each)

edit

I like adding columns to a DataFrame instead of just building an external Series.

df[‘error_for_a’] = df[a] - 70

This open possibilities and keep everything together. One could use

df[df[‘error_for_a’]>4]

Again, all() or any() can be useful but in a real life scenario, we would probably need to trig the “fault detection” when a certain number of errors are present.

If the schedule has been set ‘occupied’ at 8hAM.... maybe the first entries won’t be correct.... (any would trig an error even if the situation gets better 30minutes later). Another scenario would be a conference room where error is tiny....but as soon as there are people in it...things go bad (all() would not see that).

Solution 2:

Answer : You can iterate over columns

for col in df.columns:
    if (df[col] > 4).any(): # or .all if needed print('Zone %s does not make setpoint' % col)
    else:
        print('Zone %s is Normal' % col)

Or by defining a function and using apply

def_print(x):
    if (x > 4).any():
        print('Zone %s does not make setpoint' % x.name)
    else:
        print('Zone %s is Normal' % x.name)

df.apply(lambda x: _print(x))
# you can even do
[_print(df[col]) for col in df.columns]

Advice: maybe you would keep the result in another structure, change the function to return a boolean series that "is normal":

defis_normal(x):
    returnnot (x > 4).any()

s = df.apply(lambda x: is_normal(x))

# or directly
s = df.apply(lambda x: not (x > 4).any())

it will return a series s where index is column names of your df and values a boolean corresponding to your condition.

You can then use it to get all the Normal columns names s[s].index or the non-normal s[~s].index

Ex : I want only the normal columns of my df: df[s[s].index]

---

A complete example For the example I will use a sample df with a different condition from yours (I check if no element is lower than 4 - Normal else Does not make the setpoint )

df = pd.DataFrame(dict(a=[1,2,3],b=[2,3,4],c=[3,4,5])) # A sampleprint(df)
   a  b  c
0  1  2  3
1  2  3  4
2  3  4  5

Your use case: Print if normal or not - Solution

for col in df.columns:
if (df[col] < 4).any():
    print('Zone %s does not make setpoint' % col)
else:
    print('Zone %s is Normal' % col)

Result

Zone a is Normal
Zone b is does not make setpoint
Zone c is does not make setpoint

To illustrate my Advice : Keep the is_normal columns in a series

s = df.apply(lambda x: not (x < 4).any()) # Build the seriesprint(s)

a     True
b     False
c     False
dtype: boolprint(df[s[~s].index]) #Falsecolumns df
   b  c
023134245print(df[s[s].index]) #Truecolumns df
   a
011223