CODEWITHSUNDEEP
pythonpandas
jim jarnac
jim jarnac

Reputation: 5152

Pandas - Python: Apply() and if/then logic

I have the following DataFrames:

example  = pd.DataFrame({"dirr":[1,0,-1,-1,1,-1,0], 
                         "value": [125,130,80,8,150,251,18], 
                         "result":[np.NaN for _ in range(7)]})

I would like to perform the following operation with cummin() and cummax() on it:

example["result"].apply(lambda x : x= example["value"].cummax() if example["dirr"]==1
                           else x= example["value"].cummin() if example["dirr"]==-1
                           else x= NaN if if example["dirr"]==0
                              )

this is returning : error: invalid syntax.

Could anyone help me straightening that one up?

That would be the intended output: 

example  = pd.DataFrame({"dirr":[1,0,-1,-1,1,-1,0], 
                         "value": [125,130,80,8,150,251,18], 
                         "result":[125, NaN, 80, 8, 150, 8, NaN]})

EDIT:

So as per the answer of @su79eu7k the following function would do:

def calc(x):
    if x['dirr'] == 1:
        return np.diag(example["value"].cummax())
    elif x['dirr'] == -1:
        return np.diag(example["value"].cummin())
    else:
        return np.nan

I should be able to shove that into a lambda but still am blocked on the syntax error... which I still don't see?

example["result"]=example.apply(lambda x : np.diag(x["value"].cummax()) if x["dirr"]==1
                               else np.diag(x["value"].cummin()) if x["dirr"]==-1
                               else NaN if x["dirr"]==0
                              )

A final little nudge form you guys would be hugely appreciated.

Upvotes: 6

Views: 5962

Answers (3)

piRSquared
piRSquared

Reputation: 294218

All numpy

v = example.value.values
d = example.dirr.values
mx = np.maximum.accumulate(v)
mn = np.minimum.accumulate(v)
example['result'] = np.where(d == 1, mx, np.where(d == -1, mn, np.nan))
example

   dirr  result  value
0     1   125.0    125
1     0     NaN    130
2    -1    80.0     80
3    -1     8.0      8
4     1   150.0    150
5    -1     8.0    251
6     0     NaN     18

timing

enter image description here

Upvotes: 1

3novak
3novak

Reputation: 2544

I think it makes the most sense to use separate lines instead of an apply. If you do use the apply function, you should create a separate function and pass it through rather than making a three-line lambda.

example.loc[example['dirr'] == 1, 'result'] = \
            example.loc[example['dirr'] == 1, 'value'].cummax()
example.loc[example['dirr'] == -1, 'result'] = \
            example.loc[example['dirr'] == -1, 'value'].cummin()

>>> example
   dirr  result  value
0     1   125.0    125
1     0     NaN    130
2    -1    80.0     80
3    -1     8.0      8
4     1   150.0    150
5    -1     8.0    251
6     0     NaN     18

Alternate apply approach below.

current_max = 0
current_min = 9999

def func(df):
    global current_max
    global current_min
    if df['dirr'] == 1:
        current_max = max(current_max, df['value'])
        return current_max
    elif df['dirr'] == -1:
        current_min = min(current_min, df['value'])
        return current_min
    else:
        return np.nan

example['result'] = example.apply(func, axis=1)

Upvotes: 3

su79eu7k
su79eu7k

Reputation: 7306

I think @3novak's solution is simple and fast. But if you really want to use apply function,

def calc(x):
    if x['dirr'] == 1:
        return example["value"].cummax()
    elif x['dirr'] == -1:
        return example["value"].cummin()
    else:
        return np.nan

example['result']  = np.diag(example.apply(calc, axis=1))

print example

   dirr  result  value
0     1   125.0    125
1     0     NaN    130
2    -1    80.0     80
3    -1     8.0      8
4     1   150.0    150
5    -1     8.0    251
6     0     NaN     18

Upvotes: 2

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