I have a messy dataframe where I am trying to "flag" the rows that contain a certain number in the ids column. The values in this column represent an inclusive range: for example, "row 4" contains the following numbers:
2409,2410,2411,2412,2413,2414,2377,2378,1478,1479,1480,1481,1482,1483,1484 And in "row 0" and "row 1" the range for one of the sets is backwards (1931,1930,1929)
If I want to know which rows have sets that contain "2340" and "1930" for example, how would I do this? I think a loop is needed, sometimes will need to query more than just two numbers. Using Python 3.8.
Example Dataframe
x = ['1331:1332,1552:1551,1931:1928,1965:1973,1831:1811,1927:1920',
'1331:1332,1552:1551,1931:1929,180:178,1966:1973,1831:1811,1927:1920',
'2340:2341,1142:1143,1594:1593,1597:1596,1310,1311',
'2339:2341,1142:1143,1594:1593,1597:1596,1310:1318,1977:1974',
'2409:2414,2377:2378,1478:1484',
'2474:2476',
]
y = [6.48,7.02,7.02,6.55,5.99,6.39,]
df = pd.DataFrame(list(zip(x, y)), columns =['ids', 'val'])
display(df)
Desired Output Dataframe
I would write a function that perform 2 steps:
Given the ids_string that contains the range of ids, list all the ids as ids_num_list
Check if the query_id is in the ids_num_list
def check_num_in_ids_string(ids_string, query_id):
# Convert ids_string to ids_num_list
ids_range_list = ids_string.split(',')
ids_num_list = set()
for ids_range in ids_range_list:
if ':' in ids_range:
lower, upper = sorted(ids_range.split(":"))
num_list = list(range(int(lower), int(upper)+ 1))
ids_num_list.update(num_list)
else:
ids_num_list.add(int(ids_range))
# Check if query number is in the list
if int(query_id) in ids_num_list:
return 1
else:
return 0
# Example usage
query_id_list = ['2340', '1930']
for query_id in query_id_list:
df[f'n{query_id}'] = (
df['ids']
.apply(lambda x : check_num_in_ids_string(x, query_id))
)
which returns you what you require:
ids val n2340 n1930
0 1331:1332,1552:1551,1931:1928,1965:1973,1831:1... 6.48 0 1
1 1331:1332,1552:1551,1931:1929,180:178,1966:197... 7.02 0 1
2 2340:2341,1142:1143,1594:1593,1597:1596,1310,1311 7.02 1 0
3 2339:2341,1142:1143,1594:1593,1597:1596,1310:1... 6.55 1 0
4 2409:2414,2377:2378,1478:1484 5.99 0 0
5 2474:2476 6.39 0 0
Related
I have one dataframe with two columns , A and B . first i need to make empty bins with step 1 from 1 to 11 , (1,2),(2,3)....(10,11). then check from original dataframe if column B value greater than 3 then get value of column 'A' 2 rows before when column B is greater than 3.
Here is example dataframe :
df=pd.DataFrame({'A':[1,8.5,5.2,7,8,9,0,4,5,6],'B':[1,2,2,2,3.1,3.2,3,2,1,2]})
Required output 1:
df_out1=pd.DataFrame({'Value_A':[8.5,5.2]})
Required_output_2:
df_output2:
Bins count
(1 2) 0
(2,3) 0
(3,4) 0
(4,5) 0
(5,6) 1
(6,7) 0
（7,8） 0
（8,9） 1
（9,10） 0
（10,11） 0
You can index on a shifted series to get the two rows before 'A' satisfies some condition like
out1 = df['A'].shift(3)[df['B'] > 3]
The thing you want to do with the bins is known as a histogram. You can easily do this with numpy like
count, bin_edges = np.histogram(out1, bins=[i for i in range(1, 12)])
out2 = pd.DataFrame({'bin_lo': bin_edges[:-1], 'bin_hi': bin_edges[1:], 'count': count})
Here 'bin_lo' and 'bin_hi' are the lower and upper bounds of the bins.
Good Afternoon!
I have a pandas dataframe with an index and a count.
dictionary = {1:5,2:10,4:3,5:2}
df = pd.DataFrame.from_dict(dictionary , orient = 'index' , columns = ['count'])
What I want to do is check from df.index.min() to df.index.max() that the index increment is 1. If a value is missing like in my case the 3 is missing then I want to add 3 to the index with a 0 in the count.
The output will look like the below df2 but done in a programmatic fashion so I can use it on a much bigger dataframe.
RESULTS EXAMPLE DF:
dictionary2 = {1:5,2:10,3:0,4:3,5:2}
df2 = pd.DataFrame.from_dict(dictionary2 , orient = 'index' , columns = ['count'])
Thank you much!!!
Ensure the index is sorted:
df = df.sort_index()
Create an array that starts from the minimum index to the maximum index
complete_array = np.arange(df.index.min(), df.index.max() + 1)
Reindex, fill the null value with 0, and optionally change the dtype to Pandas Int:
df.reindex(complete_array, fill_value=0).astype("Int16")
count
1 5
2 10
3 0
4 3
5 2
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 DataFrame that looks like:
Where I have YEAR and RACEETHN as a multiindex. I want to to count the number of "1" values (note, the data are not only 0 and 1 so I cannot do a sum) for each YEAR and RACEETHN combination for each column variable.
I am able to count where value = 1 for each column by doing this:
(df_3.ACSUPPSV == 1).sum()
(df_3.PSEDSUPPSV == 1).sum()
I want to do this with groupby, but am unable to get it to work. I've tried the following code to test if I could do it on a single column 'ACSUPPSV' and it did no work:
df.groupby(['YEAR', 'RACEETHN']).loc[df.ACSUPPSV == 1, 'ACSUPPSV'].count()
I exported the data to excel and was able to calculate this with a quick "COUNTIF" formula, but I know there must be a way to do this in pandas - the results from excel look like:
Would appreciate if someone had a better way to do this than export to Excel! :)
I think you need agg with custom function for count 1 only:
df_3 = pd.DataFrame({'ACSUPPSV':[1,1,1,1,0,1],
'PSEDSUPPSV':[1,1,0,1,0,0],
'BUDGETSV':[1,0,1,1,1,0],
'YEAR':[2000,2000,2001,2000,2000,2000],
'RACEETHN':list('aaabbb')}).set_index(['YEAR','RACEETHN'])
print (df_3)
ACSUPPSV BUDGETSV PSEDSUPPSV
YEAR RACEETHN
2000 a 1 1 1
a 1 0 1
2001 a 1 1 0
2000 b 1 1 1
b 0 1 0
b 1 0 0
df2 = df_3.groupby(['YEAR', 'RACEETHN']).agg(lambda x: (x == 1).sum())
print (df2)
ACSUPPSV BUDGETSV PSEDSUPPSV
YEAR RACEETHN
2000 a 2 1 2
b 2 2 1
2001 a 1 1 0
Old answer:
df_3[((df_3.ACSUPPSV == 1) & (df_3.PSEDSUPPSV == 1))].groupby(['YEAR', 'RACEETHN']).size()
df_3.query('ACSUPPSV == 1 & PSEDSUPPSV == 1').groupby(['YEAR', 'RACEETHN']).size()
More general:
cols = ['ACSUPPSV','PSEDSUPPSV']
df_3[(df_3[cols] == 1).all(axis=1)].groupby(['YEAR', 'RACEETHN']).size()
For all columns:
df_3[(df_3 == 1).all(axis=1)].groupby(['YEAR', 'RACEETHN']).size()
EDIT:
Or maybe need:
df_3.groupby(['YEAR', 'RACEETHN']).agg(lambda x: (x == 1).sum())