I want to select a row based on a condition and then update it in dataframe.
One solution I found is to update df based on condition, but I must repeat the condition, what is the better solution so that I get the desired row once and change it?
df.loc[condition, "top"] = 1
df.loc[condition, "pred_text1"] = 2
df.loc[condtion, "pred1_score"] = 3
something like:
row = df.loc[condition]
row["top"] = 1
row["pred_text1"] = 2
row["pred1_score"] = 3
Extract the boolean mask and set it as a variable.
m = condition
df.loc[m, 'top'] = 1
df.loc[m, 'pred_text1'] = 2
df.loc[m, 'pred1_score'] = 3
but the shortest way is:
df.loc[condition, ['top', 'pred_text1', 'pred_score']] = [1, 2, 3]
Update
Wasn't it possible to retrieve the index of row and then update it by that index?
idx = df[condition].idx
df.loc[idx, 'top'] = 1
df.loc[idx, 'pred_text1'] = 2
df.loc[idx, 'pred1_score'] = 3
Related
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'm trying to get a subset of my data whenever there is consecutive occurrence of an two events in that order. The event is time-stamped. So every time there are continuous 2's and then continuous 3's, I want to subset that to a dataframe and append it to a dictionary. The following code does that but I have to apply this to a very large dataframe of more than 20 mil obs. This is extremely slow using iterrows. How can I make this fast?
df = pd.DataFrame({'Date': [101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122],
'Event': [1,1,2,2,2,3,3,1,3,2,2,3,1,2,3,2,3,2,2,3,3,3]})
dfb = pd.DataFrame(columns = df.columns)
C = {}
f1 = 0
for index, row in df.iterrows():
if ((row['Event'] == 2) & (3 not in dfb['Event'].values)):
dfb = dfb.append(row)
f1 =1
elif ((row['Event'] == 3) & (f1 == 1)):
dfb = dfb.append(row)
elif 3 in dfb['Event'].values:
f1 = 0
C[str(dfb.iloc[0,0])] = dfb
del dfb
dfb = pd.DataFrame(columns = df.columns)
if row['Event'] == 2:
dfb = dfb.append(row)
f1 =1
else:
f1=0
del dfb
dfb = pd.DataFrame(columns = df.columns)
Edit: The desired output is basically a dictionary of the subsets shown in the imagehttps://i.stack.imgur.com/ClWZs.png
If you want to accerlate, you should vectorize your code. You could try it like this (df is the same with your code):
vec = df.copy()
vec['Event_y'] = vec['Event'].shift(1).fillna(0).astype(int)
vec['Same_Flag'] = float('nan')
vec.Same_Flag.loc[(vec['Event_y'] == vec['Event']) & (vec['Event'] != 1)] = 1
vec.dropna(inplace=True)
vec.loc[:, ('Date', 'Event')]
Output is:
Date Event
3 104 2
4 105 2
6 107 3
10 111 2
18 119 2
20 121 3
21 122 3
I think that's close to what you need. You could improve based on that.
I'm not understand why date 104, 105, 107 are not counted.
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 a df that contains ids and timestamps.
I was looking to group by the id and then a condition on the timestamp in the two rows.
Something like if timestamp_col1 > timestamp_col1 for the second row then 1 else 2
Basically grouping the ids and an if statement to give a value of 1 if the first row timestamp is < than the second and 2 if the second row timestamp is < then the first
Updated output below where last two values should be 2
Use to_timedelta for converting times, then aggregate difference between first and last value and compare by gt (>), last map with numpy.where for assign new column:
df = pd.DataFrame({
'ID Code': ['a','a','b','b'],
'Time Created': ['21:25:27','21:12:09','21:12:00','21:12:40']
})
df['Time Created'] = pd.to_timedelta(df['Time Created'])
mask = df.groupby('ID Code')['Time Created'].agg(lambda x: x.iat[0] < x.iat[-1])
print (mask)
ID Code
a True
b False
Name: Time Created, dtype: bool
df['new'] = np.where(df['ID Code'].map(mask), 1, 2)
print (df)
ID Code Time Created new
0 a 21:25:27 2
1 a 21:12:09 2
2 b 21:12:00 1
3 b 21:12:40 1
Another solution with transform for return aggregate value to new column, here boolean mask:
df['Time Created'] = pd.to_timedelta(df['Time Created'])
mask = (df.groupby('ID Code')['Time Created'].transform(lambda x: x.iat[0] > x.iat[-1]))
print (mask)
0 True
1 True
2 False
3 False
Name: Time Created, dtype: bool
df['new'] = np.where(mask, 2, 1)
print (df)
ID Code Time Created new
0 a 21:25:27 2
1 a 21:12:09 2
2 b 21:12:00 1
3 b 21:12:40 1
I want to create a row number series - but not override my date index.
I can do it with a loop but I think there must be an easier way?
_cnt = [ ]
for i in range ( len ( df ) ):
_cnt.append ( i )
df[ 'row' ] = _cnt
Thanks.
Probably the easiest way:
df['row'] = range(len(df))
>>> df
0 1
0 0.444965 0.993382
1 0.001578 0.174628
2 0.663239 0.072992
3 0.664612 0.291361
4 0.486449 0.528354
>>> df['row'] = range(len(df))
>>> df
0 1 row
0 0.444965 0.993382 0
1 0.001578 0.174628 1
2 0.663239 0.072992 2
3 0.664612 0.291361 3
4 0.486449 0.528354 4