I am trying to create a new column in Pandas dataframe. If the other two date columns in my dataframe share the same month, then this new column should have 1 as a value, otherwise 0. Also, I need to check that ids match my other list of ids that I have saved previously in another place and mark those only with 1. I have some code but it is useless since I am dealing with almost a billion of rows.
my_list_of_ids = df[df.bool_column == 1].id.values
def my_func(date1, date2):
for id_ in df.id:
if id_ in my_list_of_ids:
if date1.month == date2.month:
my_var = 1
else:
my_var = 0
else:
my_var = 0
return my_var
df["new_column"] = df.progress_apply(lambda x: my_func(x['date1'], x['date2']), axis=1)
Been waiting for 30 minutes and still 0%. Any help is appreciated.
UPDATE (adding an example):
id | date1 | date2 | bool_column | new_column |
id1 2019-02-13 2019-04-11 1 0
id1 2019-03-15 2019-04-11 0 0
id1 2019-04-23 2019-04-11 0 1
id2 2019-08-22 2019-08-11 1 1
id2 ....
id3 2019-09-01 2019-09-30 1 1
.
.
.
What I need to do is save the ids that are 1 in my bool_column, then I am looping through all of the ids in my dataframe and checking if they are in the previously created list (= 1). Then I want to compare month and the year of date1 and date2 columns and if they are the same, create a new_column with a value 1 where they mach, otherwise, 0.
The pandas way to do this is
mask = ((df['date1'].month == df['date2'].month) & (df['id'].isin(my_list_of_ids)))
df['new_column'] = mask.replace({False: 0, True: 1})
Since you have a large data-set, this will take time, but should be faster than using apply
The best way to deal with the month match is to use vectorization in pandas and do this:
new_column = (df.date1.dt.month == df.date2.dt.month).astype(int)
That is, avoid using apply() over the DataFrame (which will probably be iterative) and take advantage of the underlying numpy vectorization. The gateway to such functionality is almost always in families of Series functions and properties, like the dt family for dates, str family for strings, and so forth.
Luckily, you have pre-computed the id_list membership in your bool_column, so to add membership as a criterion, just do this:
new_column = ((df.date1.dt.month == df.date2.dt.month) & df.bool_column).astype(int)
Once again, the & of two Series takes advantage of vectorization. You stay inside boolean space till the end, then cast to int with astype(int). Reviewing your code, it occurs to me that the iterative checking of your id_list may be the real performance hit here, even more so than the DataFrame.apply(). Whatever you do, avoid at all costs iterating your id_list at each row, since you already have a vector denoting membership in your bool_column.
By the way I believe there's a tiny error in your example data, the new_column value for your third row should be 0, since your bool_column value there is 0.
Related
I have a problem where I would like to count the number of times the current value has not changed in a dataframe over rolling periods.
For example:
df = pd.DataFrame({'col':list('aaaabbab')})
would somehow give output of
0
1
2
3
0
1
0
0
I have been trying something along the following
df['col'] = df['col'] == df['col'].shift(1)
df.rolling(window=3).sum().reset_index(drop=True, level=0)
I have added in the rolling as I will want to look at the full data set in terms of rolling periods but even without having it over rolling periods I can not quite figure out the logic.
I am not sure if I am missing something simple or this may not be possible using shift
You need to generate a grouper for the change in values. For this compare each value with the previous one and apply a cumsum. This gives you groups in the itertools.groupby style ([1, 1, 1, 1, 2, 2, 3, 4]), finally group and apply a cumcount.
df['count'] = (df.groupby(df['col'].ne(df['col'].shift()).cumsum())
.cumcount()
)
output:
col count
0 a 0
1 a 1
2 a 2
3 a 3
4 b 0
5 b 1
6 a 0
7 b 0
edit: for fun here is a solution using itertools (much faster):
from itertools import groupby, chain
df['count'] = list(chain(*(list(range(len(list(g))))
for _,g in groupby(df['col']))))
NB. this runs much faster (88 µs vs 707 µs on the provided example)
I can't comment so just to add some more to #mozway answer.
My goal was to count consecutives value for an entire huge dataframe effectively.
The pb I encounter is that by construction
np.nan == np.nan
will return False so you could have a whole column full of only NaN and yet the counter will be at 0.
A simple workaround would be to replace all NaN in your df by a value not already in it.
For instance in the case of a float dataset you could do
df.fillna('NA')
which will work but by changing the dtype of your columns to Object the following code will be much slower (20x on my set up).
I would rather advised something like :
all_values = list(np.unique(np.array(df)))
all_values = [a for a in all_values if a==a]
unik_val = min(all_values)-1
temp = df.fillna(unik_val).copy()
from itertools import groupby, chain
for col in temp.columns:
temp[col] = list(chain(*(list(range(len(list(g))))
for _,g in groupby(temp[col]))))
count_df
Given a dataframe
data = [['Bob','25'],['Alice','46'],['Alice','47'],['Charlie','19'],
['Charlie','19'],['Charlie','19'],['Doug','23'],['Doug','35'],['Doug','35.5']]
df = pd.DataFrame(data, columns = ['Customer','Sequence'])
Calculate the following:
First Sequence in each group is assigned a GroupID of 1.
Compare first Sequence to subsequent Sequence values in each group.
If difference is greater than .5, increment GroupID.
If GroupID was incremented, instead of comparing subsequent values to the first, use the current Sequence.
In the desired results table below...
Bob only has 1 record so the GroupID is 1.
Alice has 2 records and the difference between the two Sequence values (46 & 47) is greater than .5 so the GroupID is incremented.
Charlie's Sequence values are all the same, so all records get GroupID 1.
For Doug, the difference between the first two Sequence values (23 & 35) is greater than .5, so the GroupID for the second Sequence becomes 2. Now, since the GroupID was incremented, I want to compare the next value of 35.5 to 35, not 23, which means the last two rows share the same GroupID.
Desired results:
CustomerID
Sequence
GroupID
Bob
25
1
Alice
46
1
Alice
47
2
Charlie
19
1
Charlie
19
1
Charlie
19
1
Doug
23
1
Doug
35
2
Doug
35.5
2
My implementation:
# generate unique ID based on each customers Sequence
df['EventID'] = df.groupby('Customer')[
'Sequence'].transform(lambda x: pd.factorize(x)[0]) + 1
# impute first Sequence for each customer for comparison
df['FirstSeq'] = np.where(
df['EventID'] == 1, df['Sequence'], np.nan
)
# groupby and fill first Sequence forward
df['FirstSeq'] = df.groupby('Customer')[
'FirstSeq'].transform(lambda v: v.ffill())
# get difference of first Sequence and all others
df['FirstSeqDiff'] = abs(df['FirstSeq'] - df['Sequence'])
# create unique GroupID based on Sequence difference from first Sequence
df["GroupID"] = np.cumsum(df.FirstSeqDiff > 0.5) + 1
The above works for cases like Bob, Alice and Charlie but not Doug because it is always comparing to the first Sequence. How can I modify the code to change the compared Sequence value if the GroupID is incremented?
EDIT:
The dataframe will always be sorted by Customer and Sequence. I guess a better way to explain my goal is to assign a unique ID to all Sequence values whose difference are .5 or less, grouping by Customer.
The code has errors -> add df = df.astype({'Customer':str,'Sequence':np.float64}) would fix it. But still you cannot get what you want with this design. Try to define your own lambda function myfunc, which solves your problem directly:
data = [['Bob','25'],['Alice','46'],['Alice','47'],['Charlie','19'],
['Charlie','19'],['Charlie','19'],['Doug','23'],['Doug','35'],['Doug','35.5']]
df = pd.DataFrame(data, columns = ['Customer','Sequence'])
df = df.astype({'Customer':str,'Sequence':np.float64})
def myfunc(series):
ret = []
series = series.sort_values().values
for i,val in enumerate(series):
if i==0:
ret.append(1)
else:
ret.append(ret[-1]+(series[i]-series[i-1]>0.5))
return ret
df['EventID'] = df.groupby('Customer')[
'Sequence'].transform(lambda x: myfunc(x))
print (df)
Happy coding my friend.
I have the following df,
A id
[ObjectId('5abb6fab81c0')] 0
[ObjectId('5abb6fab81c3'),ObjectId('5abb6fab81c4')] 1
[ObjectId('5abb6fab81c2'),ObjectId('5abb6fab81c1')] 2
I like to flatten each list in A, and assign its corresponding id to each element in the list like,
A id
ObjectId('5abb6fab81c0') 0
ObjectId('5abb6fab81c3') 1
ObjectId('5abb6fab81c4') 1
ObjectId('5abb6fab81c2') 2
ObjectId('5abb6fab81c1') 2
I think the comment is coming from this question ? you can using my original post or this one
df.set_index('id').A.apply(pd.Series).stack().reset_index().drop('level_1',1)
Out[497]:
id 0
0 0 1.0
1 1 2.0
2 1 3.0
3 1 4.0
4 2 5.0
5 2 6.0
Or
pd.DataFrame({'id':df.id.repeat(df.A.str.len()),'A':df.A.sum()})
Out[498]:
A id
0 1 0
1 2 1
1 3 1
1 4 1
2 5 2
2 6 2
This probably isn't the most elegant solution, but it works. The idea here is to loop through df (which is why this is likely an inefficient solution), and then loop through each list in column A, appending each item and the id to new lists. Those two new lists are then turned into a new DataFrame.
a_list = []
id_list = []
for index, a, i in df.itertuples():
for item in a:
a_list.append(item)
id_list.append(i)
df1 = pd.DataFrame(list(zip(alist, idlist)), columns=['A', 'id'])
As I said, inelegant, but it gets the job done. There's probably at least one better way to optimize this, but hopefully it gets you moving forward.
EDIT (April 2, 2018)
I had the thought to run a timing comparison between mine and Wen's code, simply out of curiosity. The two variables are the length of column A, and the length of the list entries in column A. I ran a bunch of test cases, iterating by orders of magnitude each time. For example, I started with A length = 10 and ran through to 1,000,000, at each step iterating through randomized A entry list lengths of 1-10, 1-100 ... 1-1,000,000. I found the following:
Overall, my code is noticeably faster (especially at increasing A lengths) as long as the list lengths are less than ~1,000. As soon as the randomized list length hits the ~1,000 barrier, Wen's code takes over in speed. This was a huge surprise to me! I fully expected my code to lose every time.
Length of column A generally doesn't matter - it simply increases the overall execution time linearly. The only case in which it changed the results was for A length = 10. In that case, no matter the list length, my code ran faster (also strange to me).
Conclusion: If the list entries in A are on the order of a few hundred elements (or less) long, my code is the way to go. But if you're working with huge data sets, use Wen's! Also worth noting that as you hit the 1,000,000 barrier, both methods slow down drastically. I'm using a fairly powerful computer, and each were taking minutes by the end (it actually crashed on the A length = 1,000,000 and list length = 1,000,000 case).
Flattening and unflattening can be done using this function
def flatten(df, col):
col_flat = pd.DataFrame([[i, x] for i, y in df[col].apply(list).iteritems() for x in y], columns=['I', col])
col_flat = col_flat.set_index('I')
df = df.drop(col, 1)
df = df.merge(col_flat, left_index=True, right_index=True)
return df
Unflattening:
def unflatten(flat_df, col):
flat_df.groupby(level=0).agg({**{c:'first' for c in flat_df.columns}, col: list})
After unflattening we get the same dataframe except column order:
(df.sort_index(axis=1) == unflatten(flatten(df)).sort_index(axis=1)).all().all()
>> True
To create unique index you can call reset_index after flattening
In the flowing data frame in Pandas, I want to extract columns corresponding dates between '03/01' and '06/01'. I don't want to use the index at all, as my input would be a start and end dates. How could I do so ?
A B
0 01/01 56
1 02/01 54
2 03/01 66
3 04/01 77
4 05/01 66
5 06/01 72
6 07/01 132
7 08/01 127
First create a list of the dates you need using daterange. I'm adding the year 2000 since you need to supply a year for this to work, im then cutting it off to get the desired strings. In real life you might want to pay attention to the actual year due to things like leap-days.
date_start = '03/01'
date_end = '06/01'
dates = [x.strftime('%m/%d') for x in pd.date_range('2000/{}'.format(date_start),
'2000/{}'.format(date_end), freq='D')]
dates is now equal to:
['03/01',
'03/02',
'03/03',
'03/04',
.....
'05/29',
'05/30',
'05/31',
'06/01']
Then simply use the isin argument and you are done
df = df.loc[df.A.isin(dates)]
df
If your columns is a datetime column I guess you can skip the strftime part in th list comprehension to get the right result.
You are welcome to use boolean masking, i.e.:
df[(df.A >= start_date) && (df.A <= end_date)]
Inside the bracket is a boolean array of True and False. Only rows that fulfill your given condition (evaluates to True) will be returned. This is a great tool to have and it works well with pandas and numpy.
I have time series data with a column which can take a value A, B, or C.
An example of my data looks like this:
date,category
2017-01-01,A
2017-01-15,B
2017-01-20,A
2017-02-02,C
2017-02-03,A
2017-02-05,C
2017-02-08,C
I want to group my data by month and store both the sum of the count of A and the count of B in column a_or_b_count and the count of C in c_count.
I've tried several things, but the closest I've been able to do is to preprocess the data with the following function:
def preprocess(df):
# Remove everything more granular than day by splitting the stringified version of the date.
df['date'] = pd.to_datetime(df['date'].apply(lambda t: t.replace('\ufeff', '')), format="%Y-%m-%d")
# Set the time column as the index and drop redundant time column now that time is indexed. Do this op in-place.
df = df.set_index(df.date)
df.drop('date', inplace=True, axis=1)
# Group all events by (year, month) and count category by values.
counted_events = df.groupby([(df.index.year), (df.index.month)], as_index=True).category.value_counts()
counted_events.index.names = ["year", "month", "category"]
return counted_events
which gives me the following:
year month category
2017 1 A 2
B 1
2 C 3
A 1
The process to sum up all A's and B's would be quite manual since category becomes a part of the index in this case.
I'm an absolute pandas menace, so I'm likely making this much harder than it actually is. Can anyone give tips for how to achieve this grouping in pandas?
I tried this so posting though I like #Scott Boston's solution better as I combined A and B values earlier.
df.date = pd.to_datetime(df.date, format = '%Y-%m-%d')
df.loc[(df.category == 'A')|(df.category == 'B'), 'category'] = 'AB'
new_df = df.groupby([df.date.dt.year,df.date.dt.month]).category.value_counts().unstack().fillna(0)
new_df.columns = ['a_or_b_count', 'c_count']
new_df.index.names = ['Year', 'Month']
a_or_b_count c_count
Year Month
2017 1 3.0 0.0
2 1.0 3.0