I have one dataframe , i want to extract 2 rows before flag change from 0 to one and get row where value 'B' is minimum , also extract two rows after flag 1 and get row with minimum value of 'B'
df=pd.DataFrame({'A':[1,3,4,7,8,11,1,15,20,15,16,87],
'B':[1,3,4,6,8,11,1,19,20,15,16,87],
'flag':[0,0,0,0,1,1,1,0,0,0,0,0]})
df_out=pd.DataFrame({'A':[4,1],
'B':[4,1],
'flag':[0,1]})
To find indices of both rows of interest, run:
ind1 = df[df.flag.shift(-1).eq(0) & df.flag.shift(-2).eq(1)].index[0]
ind2 = df[df.index > ind1].B.idxmin()
For your data sample the result is 2 and 6.
Then, to retrieve rows with these indices, run:
df.loc[[ind1, ind2]]
The result is:
A B flag
2 4 4 0
6 1 1 1
I have a dataframe which has 500K rows and 7 columns for days and include start and end day.
I search a value(like equal 0) in range(startDay, endDay)
Such as, for id_1, startDay=1, and endDay=7, so, I should seek a value D1 to D7 columns.
For id_2, startDay=4, and endDay=7, so, I should seek a value D4 to D7 columns.
However, I couldn't seek different column range successfully.
Above-mentioned,
if startDay > endDay, I should see "-999"
else, I need to find first zero (consider the day range) and such as for id_3's, first zero in D2 column(day 2). And starDay of id_3 is 1. And I want to see, 2-1=1 (D2 - StartDay)
if I cannot find 0, I want to see "8"
Here is my data;
data = {
'D1':[0,1,1,0,1,1,0,0,0,1],
'D2':[2,0,0,1,2,2,1,2,0,4],
'D3':[0,0,1,0,1,1,1,0,1,0],
'D4':[3,3,3,1,3,2,3,0,3,3],
'D5':[0,0,3,3,4,0,4,2,3,1],
'D6':[2,1,1,0,3,2,1,2,2,1],
'D7':[2,3,0,0,3,1,3,2,1,3],
'startDay':[1,4,1,1,3,3,2,2,5,2],
'endDay':[7,7,6,7,7,7,2,1,7,6]
}
data_idx = ['id_1','id_2','id_3','id_4','id_5',
'id_6','id_7','id_8','id_9','id_10']
df = pd.DataFrame(data, index=data_idx)
What I want to see;
df_need = pd.DataFrame([0,1,1,0,8,2,8,-999,8,1], index=data_idx)
You can create boolean array to check in each row which 'Dx' column(s) are above 'startDay' and below 'endDay' and the value is equal to 0. For the first two conditions, you can use np.ufunc.outer with the ufunc being np.less_equal and np.greater_equal such as:
import numpy as np
arr_bool = ( np.less_equal.outer(df.startDay, range(1,8)) # which columns Dx is above startDay
& np.greater_equal.outer(df.endDay, range(1,8)) # which columns Dx is under endDay
& (df.filter(regex='D[0-9]').values == 0)) #which value of the columns Dx are 0
Then you can use np.argmax to find the first True per row. By adding 1 and removing 'startDay', you get the values you are looking for. Then you need to look for the other conditions with np.select to replace values by -999 if df.startDay >= df.endDay or 8 if no True in the row of arr_bool such as:
df_need = pd.DataFrame( (np.argmax(arr_bool , axis=1) + 1 - df.startDay).values,
index=data_idx, columns=['need'])
df_need.need= np.select( condlist = [df.startDay >= df.endDay, ~arr_bool.any(axis=1)],
choicelist = [ -999, 8],
default = df_need.need)
print (df_need)
need
id_1 0
id_2 1
id_3 1
id_4 0
id_5 8
id_6 2
id_7 -999
id_8 -999
id_9 8
id_10 1
One note: to get -999 for id_7, I used the condition df.startDay >= df.endDay in np.select and not df.startDay > df.endDay like in your question, but you can cahnge to strict comparison, you get 8 instead of -999 in this case.
I have a csv file named namelist.csv, it includes:
Index String Size Name
1 AAA123000DDD 10 One
2 AAA123DDDQQQ 20 One
3 AAA123000DDD 25 One
4 AAA123D 20 One
5 ABA 15 One
6 FFFrrrSSSBBB 60 Two
7 FFFrrrSSSBBB 30 Two
8 FFFrrrSS 50 Two
9 AAA12 70 Two
I want to compare row in column String of each name group: if the string in each row is match or is substring of all above rows then remove the previous rows and sum the value of Size column to the value of subtring row.
Example: i take row 3rd: AAA123000DDD, i compare it to 2 row 1st and 2nd, it see that it is a match with 1st row, it will remove the 1st row then sum value of the 1st row column Size to the 3rd row column Size .
then the table will be like:
Index String Size Name
2 AAA123DDDQQQ 20 One
3 AAA123000DDD 35 One
4 AAA123D 20 One
...
the final result will be:
Index String Size Name
3 AAA123000DDD 35 One
4 AAA123D 40 One
5 ABA 15 One
8 FFFrrrSS 140 Two
9 AAA12 70 Two
i think of using groupby of pandas to group all Name column, but i don't know how to apply the comparison of String column and sum of Size column.
I am new to Python so any help I will very appreciate.
Assuming Name is distinct with String, here's how you would do the aggregation. I kept Name so that it also shows in the final DataFrame.
df_group = df.groupby(['String', 'Name'])['Size'].sum().reset_index()
Edit:
To match the substrings (and using the example above that it appears that a substring will not match with multiple strings), you can make a mapping of substrings to full strings and then group by the full string column as before:
all_strings = set(df['Strings'])
substring_dict = dict()
for row in df.itertuples():
for item in all_strings:
if row.String in item:
substring_dict[row.String] = item
def match_substring(x):
return substring_dict[x]
df['full_strings'] = df.String.apply(match_substring)
df_group = df.groupby(['full_strings', 'Name'])['Size'].sum().reset_index()
Data Looks Like:
col 1 col 2 col 3 col 4
row 1 row 1 row 1 row 1
row 2 row 2 row 2 row 2
row 3 row 3 row 3 row 3
row 4 row 4 row 4 row 4
row 5 row 5 row 5 row 5
row 6 row 6 row 6 row 6
Problem: I want to partition this data, lets say row 1 and row 2 will be processed as one partition, row 3 and row 4 as another, row 5 and row 6 as another and create a JSON data merging them together with the column (column headers with data values in rows).
Output should be like:
[
{col1:row1,col2:row1:col3:row1:col4:row1},
{col1:row2,col2:row2:col3:row2:col4:row2},
{col1:row3,col2:row3:col3:row3:col4:row3},
{col1:row4,col2:row4:col3:row4:col4:row4},...
]
I tried using repartion(num) available in spark but it is not exactly partitioning as i want. therefore the JSON data generated is not valid. i had issue with why my program was taking same time for processing the data even though i was using different number of cores which can be found here and the repartition suggestion was suggested by #Patrick McGloin . The code mentioned in that problem is something i am trying to do.
Guess what you need is partitionBy. In Scala you can provide to it a custom build HashParitioner, while in Python you pass partitionFunc. There is a number of examples out there in Scala, so let me briefly explain the Python flavour.
partitionFunc expects a tuple, with first element being the key. Lets assume you organise your data in the following fashion:
(ROW_ID, (A,B,C,..)) where ROW_ID = [1,2,3,...,k]. You can always add ROW_ID and remove it afterwards.
To get a new partition every two rows:
rdd.partitionBy(numPartitions = int(rdd.count() / 2),
partitionFunc = lambda key: int(key / 2)
partitionFunc will produce a sequence 0,0,1,1,2,2,... This number will be a number of partition to which given row will belong.