This question has been asked but I didn't find the answers complete. I have a dataframe that has unnecessary values in the first row and I want to find the row index of the animals:
df = pd.DataFrame({'a':['apple','rhino','gray','horn'],
'b':['honey','elephant', 'gray','trunk'],
'c':['cheese','lion', 'beige','mane']})
a b c
0 apple honey cheese
1 rhino elephant lion
2 gray gray beige
3 horn trunk mane
ani_pat = r"rhino|zebra|lion"
That means I want to find "1" - the row index that matches the pattern. One solution I saw here was like this; applying to my problem...
def findIdx(df, pattern):
return df.apply(lambda x: x.str.match(pattern, flags=re.IGNORECASE)).values.nonzero()
animal = findIdx(df, ani_pat)
print(animal)
(array([1, 1], dtype=int64), array([0, 2], dtype=int64))
That output is a tuple of NumPy arrays. I've got the basics of NumPy and Pandas, but I'm not sure what to do with this or how it relates to the df above.
I altered that lambda expression like this:
df.apply(lambda x: x.str.match(ani_pat, flags=re.IGNORECASE))
a b c
0 False False False
1 True False True
2 False False False
3 False False False
That makes a little more sense. but still trying to get the row index of the True values. How can I do that?
We can select from the filter the DataFrame index where there are rows that have any True value in them:
idx = df.index[
df.apply(lambda x: x.str.match(ani_pat, flags=re.IGNORECASE)).any(axis=1)
]
idx:
Int64Index([1], dtype='int64')
any on axis 1 will take the boolean DataFrame and reduce it to a single dimension based on the contents of the rows.
Before any:
a b c
0 False False False
1 True False True
2 False False False
3 False False False
After any:
0 False
1 True
2 False
3 False
dtype: bool
We can then use these boolean values as a mask for index (selecting indexes which have a True value):
Int64Index([1], dtype='int64')
If needed we can use tolist to get a list instead:
idx = df.index[
df.apply(lambda x: x.str.match(ani_pat, flags=re.IGNORECASE)).any(axis=1)
].tolist()
idx:
[1]
Related
there seems to be different methods to check if a cell is not set (NaN, by checking isnull) or whether it contains an empty string or list, but what is the most pythonic way to retrieve all cells that are NaN, None, empty string/list, etc. at the same time?
So far I got:
df = df[df['colname'].isnull() or df['colname'] == None or len(df['colname']) == 0]
Cheers!
One idea is chain Series.isna with compare lengths by Series.str.len:
df = pd.DataFrame({
'a':[None,np.nan,[],'','aa', 0],
})
m = df['a'].isna() | df['a'].str.len().eq(0)
print (m)
0 True
1 True
2 True
3 True
4 False
5 False
Name: a, dtype: bool
Now I know how to check the dataframe for specific values across multiple columns. However, I cant seem to work out how to carry out an if statement based on a boolean response.
For example:
Walk directories using os.walk and read in a specific file into a dataframe.
for root, dirs, files in os.walk(main):
filters = '*specificfile.csv'
for filename in fnmatch.filter(files, filters):
df = pd.read_csv(os.path.join(root, filename),error_bad_lines=False)
Now checking that dataframe across multiple columns. The first value being the column name (column1), the next value is the specific value I am looking for in that column(banana). I am then checking another column (column2) for a specific value (green). If both of these are true I want to carry out a specific task. However if it is false I want to do something else.
so something like:
if (df['column1']=='banana') & (df['colour']=='green'):
do something
else:
do something
If you want to check if any row of the DataFrame meets your conditions you can use .any() along with your condition . Example -
if ((df['column1']=='banana') & (df['colour']=='green')).any():
Example -
In [16]: df
Out[16]:
A B
0 1 2
1 3 4
2 5 6
In [17]: ((df['A']==1) & (df['B'] == 2)).any()
Out[17]: True
This is because your condition - ((df['column1']=='banana') & (df['colour']=='green')) - returns a Series of True/False values.
This is because in pandas when you compare a series against a scalar value, it returns the result of comparing each row of that series against the scalar value and the result is a series of True/False values indicating the result of comparison of that row with the scalar value. Example -
In [19]: (df['A']==1)
Out[19]:
0 True
1 False
2 False
Name: A, dtype: bool
In [20]: (df['B'] == 2)
Out[20]:
0 True
1 False
2 False
Name: B, dtype: bool
And the & does row-wise and for the two series. Example -
In [18]: ((df['A']==1) & (df['B'] == 2))
Out[18]:
0 True
1 False
2 False
dtype: bool
Now to check if any of the values from this series is True, you can use .any() , to check if all the values in the series are True, you can use .all() .
I have problem with replacement values in a column conditional other two columns.
For example we have three columns. A, B, and C
Columns A and B are both booleans, containing True and False, and column C contains three values: "Payroll", "Social", and "Other".
When in columns A and B are True in column C we have value "Payroll".
I want to change values in column C where both column A and B are True.
I tried following code: but gives me this error "'NoneType' object has no attribute 'where'":
data1.replace({'C' : { 'Payroll', 'Social'}},inplace=True).where((data1['A'] == True) & (data1['B'] == True))
but gives me this error "'NoneType' object has no attribute 'where'":
What can be done to this problem?
I think you need all for check if all Trues per rows and then assign output by filtered DataFrame by boolean mask:
data1 = pd.DataFrame({
'C': ['Payroll','Other','Payroll','Social'],
'A': [True, True, True, False],
'B':[False, True, True, False]
})
print (data1)
A B C
0 True False Payroll
1 True True Other
2 True True Payroll
3 False False Social
m = data1[['A', 'B']].all(axis=1)
#same output as
#m = data1['A'] & data1['B']
print (m)
0 False
1 True
2 True
3 False
dtype: bool
print (data1[m])
A B C
1 True True Other
2 True True Payroll
data1[m] = data1[m].replace({'C' : { 'Payroll':'Social'}})
print (data1)
A B C
0 True False Payroll
1 True True Other
2 True True Social
3 False False Social
Well you can use apply function to do this
def change_value(dataframe):
for index, row in df.iterrows():
if row['A'] == row['B'] == True:
row['C'] = # Change to whatever value you want
else:
row ['C'] = # Change how ever you want
I'm keeping score in a True/False column when determining whether some signal is below the background level, so for example
sig bg is_below
5 3 False
5 3 False
5 3 False
2 3 True # "False positive"
4 3 False
4 3 False
0 3 True # Signal is dead and not returning from this point onwards
0 3 True
0 3 True
0 3 True
0 3 True
But as I've shown, noise may sometimes generate "false positives", and smoothing the data doesn't get rid of some big spikes, without oversmoothing smaller data. I'm sure there's a proper mathematical way, but perhaps that would be overkill in work and computational efficiency.
Instead, how do I determine the index of the first True where True appears e.g. 3 times in a row?
Okay, so I just remembered that True/False could just as easily be interpreted as 1/0, and so a rolling median, e.g.
scipy.signal.medfilt(df["is_below"], kernel_size = 5).argmax()
Would return the index of the first time encountering [False, False, True, True, True], as the median of [0, 0, 1, 1, 1] is the smallest window that returns 3 True in a row.
I don't know if there is an even better way, but given that I have 100s of datapoints in my timeseries, the returned argmax index is accurate enough for my application.
If your data is in a pandas dataframe (say called df), you can do it by creating a boolean variable b which is True at each row only when the row and previous two rows are True in df.is_below.
b = ((df.is_below == True) & (df.is_below.shift(-1) == True) & (df.is_below.shift(-2) == True))
Here, df.is_below.shift(-1) shifts the whole dataframe back by 1, so we are looking at the previous row (and similarly for shift(-2) to look at the row before the previous row).
Full code below:
import pandas as pd
# Create dataframe
df = pd.DataFrame()
sig = [5, 5, 5, 2, 4, 4, 0, 0, 0, 0, 0]
df['sig'] = sig
df['bg'] = [3] * len(sig)
df['is_below'] = df.sig < df.bg
# Find index of first consecutive three True in df.is_below
b = ((df.is_below == True) & (df.is_below.shift(-1) == True) & (df.is_below.shift(-2) == True))
idx = df.index[b][0] # first index where three Trues are in a row
I have implemented the CN2 classification algorithm, it induces rules to classify the data of the form:
IF Attribute1 = a AND Attribute4 = b THEN class = class 1
My current implementation loops through a pandas DataFrame containing the training data using the iterrows() function and returns True or False for each row if it satisfies the rule or not, however, I am aware this is a highly inefficient solution. I would like to vectorise the code, my current attempt is like so:
DataFrame = df
age prescription astigmatism tear rate
1 1 2 1
2 2 1 1
2 1 1 2
rule = {'age':[1],'prescription':[1],'astigmatism':[1,2],'tear rate':[1,2]}
df.isin(rule)
This produces:
age prescription astigmatism tear rate
True True True True
False False True True
False True True True
I have coded the rule to be a dictionary which contains a single value for target attributes and the set of all possible values for non-target attributes.
The result I would like is a single True or False for each row if the conditions of the rule are met or not and the index of the rows which evaluate to all True. Currently I can only get a DataFrame with a T/F for each value. To be concrete, in the example i have shown, I wish the result to be the index of the first row which is the only row which satisfies the rule.
I think you need check if at least one value per row is True use DataFrame.any:
mask = df.isin(rule).any(axis=1)
print (mask)
0 True
1 True
2 True
dtype: bool
Or for check if all values are Trues use DataFrame.all:
mask = df.isin(rule).all(axis=1)
print (mask)
0 True
1 False
2 False
dtype: bool
For filtering is possible use boolean indexing:
df = df[mask]