date_0 = list(pd.date_range('2017-01-01', periods=6, freq='MS'))
date_1 = list(pd.date_range('2017-01-01', periods=8, freq='MS'))
data_0 = [9, 8, 4, 0, 0, 0]
data_1 = [9, 9, 0, 0, 0, 7, 0, 0]
id_0 = [0]*6
id_1 = [1]*8
df = pd.DataFrame({'ids': id_0 + id_1, 'dates': date_0 + date_1, 'data': data_0 + data_1})
For each id (here 0 and 1) I want to know how long is the series of zeros at the end of the time frame.
For the given example, the result is id_0 = 3, id_1 = 2.
So how do I limit the timestamps, so I can run something like that:
df.groupby('ids').agg('count')
First need get all consecutive 0 with trick by compare with shifted values for not equal and cumsum.
Then count pre groups, remove first level of MultiIndex and get last values per group by drop_duplicates with keep='last':
s = df['data'].ne(df['data'].shift()).cumsum().mul(~df['data'].astype(bool))
df = (s.groupby([df['ids'], s]).size()
.reset_index(level=1, drop=True)
.reset_index(name='val')
.drop_duplicates('ids', keep='last'))
print (df)
ids val
1 0 3
4 1 2
Related
I have data like below:
df = pd.DataFrame()
df["collection_amount"] = 100, 200, 300
df["25%_coll"] = 1, 0, 1
df["75%_coll"] = 0, 1, 1
df["month"] = 4, 5, 6
I want to create a output like below:
basically if 25% is 1 then it should create a column based on month as a new column.
Please help me thank you.
This should work: do ask if something doesn't make sense
for i in range(len(df)):
if df['25%_coll'][i]==1:
df['month_%i_25%%_coll'%df.month[i]]=[df.collection_amount[i] if k==i else 0 for k in range(len(df))]
if df['75%_coll'][i]==1:
df['month_%i_75%%_coll'%df.month[i]]=[df.collection_amount[i] if k==i else 0 for k in range(len(df))]
To build the new columns you could try the following:
df2 = df.melt(id_vars=["month", "collection_amount"])
df2.loc[df2["value"].eq(0), "collection_amount"] = 0
df2["new_cols"] = "month_" + df2["month"].astype("str") + "_" + df2["variable"]
df2 = df2.pivot_table(
index="month", columns="new_cols", values="collection_amount",
fill_value=0, aggfunc="sum"
).reset_index(drop=True)
.melt() the dataframe with index columns month and collection_amount.
Set the appropriate collection_amount values to 0.
Build the new column names in column new_cols.
month collection_amount variable value new_cols
0 4 100 25%_coll 1 month_4_25%_coll
1 5 0 25%_coll 0 month_5_25%_coll
2 6 300 25%_coll 1 month_6_25%_coll
3 4 0 75%_coll 0 month_4_75%_coll
4 5 200 75%_coll 1 month_5_75%_coll
5 6 300 75%_coll 1 month_6_75%_coll
Use .pivot_table() on this dataframe to build the new columns.
The rest isn't completely clear: Either use df = pd.concat([df, df2], axis=1), or df.merge(df2, ...) to merge on month (with .reset_index() without drop=True).
Result for the sample dataframe
df = pd.DataFrame({
"collection_amount": [100, 200, 300],
"25%_coll": [1, 0, 1], "75%_coll": [0, 1, 1],
"month": [4, 5, 6]
})
is
new_cols month_4_25%_coll month_4_75%_coll month_5_25%_coll \
0 100 0 0
1 0 0 0
2 0 0 0
new_cols month_5_75%_coll month_6_25%_coll month_6_75%_coll
0 0 0 0
1 200 0 0
2 0 300 300
I have a dataframe with two integer columns that represent the start and end of a string of text. I'd like to group my rows by length of text (end - start), but with a margin of error of +- 5 characters so that something like this would happen:
start end
0 251
1 250
2 250
0 500
1 500
0 499
How would I achieve something like this?
Here is the code I am using right now
d = {'text': ["aaa", "bbb", "ccc", "ddd", "eee", "fff"],
'start': [0, 1, 0, 2, 1, 0],
'end': [250, 500, 501, 251, 249, 499]}
df = pd.DataFrame(data=d)
df = df.groupby(['start', 'end'])
I ended up solving the problem by rounding the length of my text.
df['rounded_length'] = (df['end'] - df['start']).round(-1)
df = df.groupby('rounded_length')
All my values become multiples of 10, and I can group them this way.
I have a dataframe (df) as following
id date t_slot dayofweek label
1 2021-01-01 2 0 1
1 2021-01-02 3 1 0
2 2021-01-01 4 6 1
.......
The data frame is very large(6 million rows). the t_slot is from 1 to 6 value. dayofweek is from 0-6.
I want to get the rate:
- the each id's rate about the label is 1 rate when the t_slot is 1 to 4, and dayofweek is 0-4 in the past 3 months before the date in each row.
- the each id's rate about the label is 1 rate when the t_slot is 1 to 4, and dayofweek is 0-4 in the past 3 months before the date in each row.
- the each id's rate about the label is 1 rate when the t_slot is 5 to 6, and dayofweek is 5-6 in the past 3 months before the date in each row.
- the each id's rate about the label is 1 rate when the t_slot is 5 to 6, and dayofweek is 5-6 in the past 3 months before the date in each row.
I have used loop to compute the rate, but it is very slow, do you have fast way to compute it. My code is copied as following:
def get_time_slot_rate(df):
import numpy as np
if len(df)==0:
return np.nan, np.nan, np.nan, np.nan
else:
work = df.loc[df['dayofweek']<5]
weekend = df.loc[df['dayofweek']>=5]
if len(work)==0:
work_14, work_56 = np.nan, np.nan
else:
work_14 = len(work.loc[(work['time_slot']<5)*(work['label']==1)])/len(work)
work_56 = len(work.loc[(work['time_slot']>5)*(work['label']==1)])/len(work)
if len(weekend)==0:
weekend_14, weekend_56 = np.nan, np.nan
else:
weekend_14 = len(weekend.loc[(weekend['time_slot']<5)*(weekend['label']==1)])/len(weekend)
weekend_56 = len(weekend.loc[(weekend['time_slot']>5)*(weekend['label']==1)])/len(weekend)
return work_14, work_56, weekend_14, weekend_56
import datetime as d_t
lst_id = list(df['id'])
lst_date = list(df['date'])
lst_t14_work = []
lst_t56_work = []
lst_t14_weekend = []
lst_t56_weekend = []
for i in range(len(lst_id)):
if i%100==0:
print(i)
d_date = lst_date[i]
dt = d_t.datetime.strptime(d_date, '%Y-%m-%d')
month_step = relativedelta(months=3)
pre_date = str(dt - month_step).split(' ')[0]
df_s = df.loc[(df['easy_id']==lst_easy[i])
& ((df['delivery_date']>=pre_date)
&(df['delivery_date']< d_date))].reset_index(drop=True)
work_14_rate, work_56_rate, weekend_14_rate, weekend_56_rate = get_time_slot_rate(df_s)
lst_t14_work.append(work_14_rate)
lst_t56_work.append(work_56_rate)
lst_t14_weekend.append(weekend_14_rate)
lst_t56_weekend.append(weekend_56_rate)
I could only fix your function and it's completely untested, but here we go:
Import only once by putting the imports at the top of your .py.
try/except blocks are more efficient than if/else statements.
True and False equals to 1 and 0 respectively in Python.
Don't multiply boolean selectors and use the reverse operator ~
Create the least amount of copies.
import numpy as np
def get_time_slot_rate(df):
# much faster than counting
if df.empty:
return np.nan, np.nan, np.nan, np.nan
# assuming df['label'] is either 0 or 1
df = df.loc[df['label']]
# create boolean selectors to be inverted with '~'
weekdays = df['dayofweek']<=5
slot_selector = df['time_slot']<=5
weekday_count = np.sum(weekdays)
try:
work_14 = len(df.loc[weekdays & slot_selector])/weekday_count
work_56 = len(df.loc[weekdays & ~slot_selector])/weekday_count
except ZeroDivisionError:
work_14 = work_56 = np.nan
weekend_count = np.sum(~weekdays)
try:
weekend_14 = len(df.loc[~weekdays & slot_selector])/weekend_count
weekend_56 = len(df.loc[~weekdays & ~slot_selector])/weekend_count
except ZeroDivisionError:
weekend_14 = weekend_56 = np.nan
return work_14, work_56, weekend_14, weekend_56
The rest of your script doesn't really make sense, see my comments:
for i in range(len(lst_id)):
if i%100==0:
print(i)
d_date = date[i]
# what is d_t ?
dt = d_t.datetime.strptime(d_date, '%Y-%m-%d')
month_step = relativedelta(months=3)
pre_date = str(dt - month_step).split(' ')[0]
df_s = df.loc[(df['easy_id']==lst_easy[i])
& (df['delivery_date']>=pre_date)
&(df['delivery_date']< d_date)].reset_index(drop=True)
# is it df or df_s ?
work_14_rate, work_56_rate, weekend_14_rate, weekend_56_rate = get_time_slot_rate(df)
If your date column is a datetime object than you can compare dates directly (no need for strings).
I have the following dataframe
df = pd.DataFrame([{'id':'a', 'val':1}, {'id':'b', 'val':2}, {'id':'c', 'val': 0}, {'id':'d', 'val':0}])
What I want is to replace 0's with +1 of the max value
The result I want is as follows:
df = pd.DataFrame([{'id':'a', 'val':1}, {'id':'b', 'val':2}, {'id':'c', 'val': 3}, {'id':'d', 'val':4}])
I tried the following:
for _, r in df.iterrows():
if r.val == 0:
r.val = df.val.max()+1
However, it there a one-line way to do the above
Filter only 0 rows with boolean indexing and DataFrame.loc and assign range with count Trues values of condition with add maximum value and 1, because python count from 0 in range:
df.loc[df['val'].eq(0), 'val'] = range(df['val'].eq(0).sum()) + df.val.max() + 1
print (df)
id val
0 a 1
1 b 2
2 c 3
3 d 4
I have the following code that updates the current row based on the status of the previous row:
prev_status = 0
for idx, row in df.iterrows():
if prev_status in [1, 2] and row[column_a] != 0:
row[column_b] += row[column_a]
row[column_c] = 0
row[column_d] = 0
row[column_a] = 0
prev_status = row[status]
df.loc[idx] = row
However this is very slow when running on 1GB of data. What are ways to optimize this?
Try this:
df['previous_status'] = df['status'].shift(1)
df.loc[df['previous_status'] in [1, 2] & df['column_a'] != 0, 'column_b'] += df['column_a']
df.loc[df['previous_status'] in [1, 2] & df['column_a'] != 0, 'column_c'] = 0
df.loc[df['previous_status'] in [1, 2] & df['column_a'] != 0, 'column_d'] = 0
df.loc[df['previous_status'] in [1, 2] & df['column_a'] != 0, 'column_a'] = 0
Look at using shift, e.g.
df["new_column"] = df["column_name"].shift(x)
This creates a column where the values are the values of another column shifted by x number of rows. It makes it much quicker to do vectorwise calculations on a column, rather than applying a function to every row in the DataFrame.