How to modify a data frame which looks like this:
col_name No. Missing row_number
ST_NUM 2 2,4
ST_NAME 0
OWN_OCCUPIED 3 1,3,10
NUM_BEDROOMS 2 1,4
want to convert the above data frame as:
col_name row_number
ST_NUM 2
ST_NUM 4
OWN_OCCUPIED 1
OWN_OCCUPIED 3
OWN_OCCUPIED 10
NUM_BEDROOMS 1
NUM_BEDROOMS 4
Assuming you're using pandas 0.25+, you can simply write:
df.row_number = df.row_number.str.split(",")
df.explode("row_number")[["col_name", "row_number"]]
You will then need to filter out rows corresponding to empty row_numbers, though how you do that will depend on how you represented those in the first place ("", nan, etc).
Based on this answer, you could try this :
lengths = [len(item) for item in df['row_number']]
result=pd.DataFrame( {"col_name" : np.repeat(df['col_name'].values,lengths),
"No_Missing" : np.repeat(df['No_Missing'].values,lengths),
"row_number" : np.hstack(df['row_number'])})
print(result)
col_name No_Missing row_number
0 ST_NUM 2 2.0
1 ST_NUM 2 4.0
2 OWN_OCCUPIED 3 1.0
3 OWN_OCCUPIED 3 3.0
4 OWN_OCCUPIED 3 10.0
5 NUM_BEDROOMS 2 1.0
6 NUM_BEDROOMS 2 4.0
Related
Starting from this data frame:
id
date
value
1
01.01.
2
2
01.01.
3
1
01.03.
5
2
01.03.
3
1
01.09.
5
2
01.09.
2
1
01.10.
5
2
01.10.
2
I would like to get a weekly sum of value:
id
date
value
1
01.01.
2
2
01.01.
3
1
01.03.
7
2
01.03.
6
1
01.09.
10
2
01.09.
5
1
01.10.
15
2
01.10.
8
I use this command, but it is not working:
df['value'] = df.groupby('id')['value'].rolling(7).sum()
Any ideas?
You can do groupby and apply.
df['date'] = pd.to_datetime(df['date'], format='%m.%d.')
df['value'] = (df.groupby('id', as_index=False, group_keys=False)
.apply(lambda g: g.rolling('7D', on='date')['value'].sum()))
Note that for 1900-01-10, the rolling window is 1900-01-04, 1900-01-05...1900-01-10
print(df)
id date value
0 1 1900-01-01 2.0
1 2 1900-01-01 3.0
2 1 1900-01-03 7.0
3 2 1900-01-03 6.0
4 1 1900-01-09 10.0
5 2 1900-01-09 5.0
6 1 1900-01-10 10.0
7 2 1900-01-10 4.0
I have a dataframe of locations:
df = X Y
1 1
2 1
2 1
2 2
3 3
5 5
5.5 5.5
I want to add a columns, with the distance to the previous point:
So it will be:
df = X Y Distance
1 1 0
2 1 1
2 1 0
2 2 1
3 3 2
5 5 2
5.5 5.5 1
What is the best way to do so?
You can use the pd.Series.diff method.
For instance, to compute the eulerian distance, using also np.sqrt, you would do like this:
import numpy as np
df["Distance"] = np.sqrt(df.X.diff()**2 + df.Y.diff()**2)
I have a dataframe that contains nan values in particular column. while iterating through the rows, if it come across nan(using isnan() method) then I need to change it to some other value(since I have some conditions). I tried using replace() and fillna() with limit parameter also but they are modifying whole column when they come across the first nan value? Is there any method that I can assign value to specific nan rather than changing all the values of a column?
Example: the dataframe looks like it:
points sundar cate king varun vicky john charlie target_class
1 x2 5 'cat' 4 10 3 2 1 NaN
2 x3 3 'cat' 1 2 3 1 1 NaN
3 x4 6 'lion' 8 4 3 7 1 NaN
4 x5 4 'lion' 1 1 3 1 1 NaN
5 x6 8 'cat' 10 10 9 7 1 0.0
an I have a list like
a = [1.0, 0.0]
and I expect to be like
points sundar cate king varun vicky john charlie target_class
1 x2 5 'cat' 4 10 3 2 1 1.0
2 x3 3 'cat' 1 2 3 1 1 1.0
3 x4 6 'lion' 8 4 3 7 1 1.0
4 x5 4 'lion' 1 1 3 1 1 0.0
5 x6 8 'cat' 10 10 9 7 1 0.0
I wanted to change the target_class values based on some conditions and assign values of the above list.
I believe need replace NaNs values to 1 only for indexes specified in list idx:
mask = df['target_class'].isnull()
idx = [1,2,3]
df.loc[mask, 'target_class'] = df[mask].index.isin(idx).astype(int)
print (df)
points sundar cate king varun vicky john charlie target_class
1 x2 5 'cat' 4 10 3 2 1 1.0
2 x3 3 'cat' 1 2 3 1 1 1.0
3 x4 6 'lion' 8 4 3 7 1 1.0
4 x5 4 'lion' 1 1 3 1 1 0.0
5 x6 8 'cat' 10 10 9 7 1 0.0
Or:
idx = [1,2,3]
s = pd.Series(df.index.isin(idx).astype(int), index=df.index)
df['target_class'] = df['target_class'].fillna(s)
EDIT:
From comments solution is assign values by index and columns values with DataFrame.loc:
df2.loc['x2', 'target_class'] = list1[0]
I suppose your conditions for imputing the nan values does not depend on the number of them in a column. In the code below I stored all the imputation rules in one function that receives as parameters the entire row (containing the nan) and the column you are investigating for. If you also need all the dataframe for the imputation rules, just pass it through the replace_nan function. In the example I imputate the col element with the mean values of the other columns.
import pandas as pd
import numpy as np
def replace_nan(row, col):
row[col] = row.drop(col).mean()
return row
df = pd.DataFrame(np.random.rand(5,3), columns = ['col1', 'col2', 'col3'])
col_to_impute = 'col1'
df.loc[[1, 3], col_to_impute] = np.nan
df = df.apply(lambda x: replace_nan(x, col_to_impute) if np.isnan(x[col_to_impute]) else x, axis=1)
The only thing that you should do is making the right assignation. That is, make an assignation in the rows that contain nulls.
Example dataset:
,event_id,type,timestamp,label
0,asd12e,click,12322232,0.0
1,asj123,click,212312312,0.0
2,asd321,touch,12312323,0.0
3,asdas3,click,33332233,
4,sdsaa3,touch,33211333,
Note: The last two rows contains nulls in column: 'label'. Then, we load the dataset:
df = pd.read_csv('dataset.csv')
Now, we make the appropiate condition:
cond = df['label'].isnull()
Now, we make the assignation over these rows (I don't know the logical of assignation. Therefore I assign 1 value to NaN's):
df1.loc[cond,'label'] = 1
There are another more accurate approaches. fillna() method could be used. You should provide the logical in order to help you.
I have a dataframe
L1_1 L1_2 L3_1 L2_1 L2_2 L1_3 L2_3 L3_2 ....
3 5 1 5 7 2 3 2
4 2 4 1 4 1 4 2
I need to find the mean all "L1" , then all "L2" , and then all "L3 "
I tried
data["Mean"]=data.mean(axis=1)
that give me the sum across all "L1, L2 and L3" together
I also tried
data[['L1_1','L1_2','L1_3','Mean']].head()
but I have L1_1 to L1_20
so a loop sounds good. However, I cannot get a loop to work.
for i in range(1,21):
c = "'L1_" + i + "'," + c
Is a loop a good way to go here? or Is there a better?
If a loop is the way to go, How do you get a loop to work in a data frame?
Use groupby by columns (axis=1) with custom function of splitted values:
df1 = df.groupby(lambda x: x.split('_')[0], axis=1).mean()
#another solution
#df1 = df.groupby(df.columns.str.split('_').str[0], axis=1).mean()
print (df1)
L1 L2 L3
0 3.333333 5.0 1.5
1 2.333333 3.0 3.0
If want add nex columns to original df add join with add_suffix if want also rename columns names:
df = df.join(df1.add_suffix('_mean'))
print (df)
L1_1 L1_2 L3_1 L2_1 L2_2 L1_3 L2_3 L3_2 L1_mean L2_mean L3_mean
0 3 5 1 5 7 2 3 2 3.333333 5.0 1.5
1 4 2 4 1 4 1 4 2 2.333333 3.0 3.0
I am new to Python and Pandas , can someone help me with below report.
I want to report difference of N columns and create new columns with difference value, is it possible to make it dynamic as I have more than 30 columns. (Columns are fixed numbers, rows values can change)
A and B can be Alpha numeric
Use join with sub for difference of DataFrames:
#if columns are strings, first cast it
df1 = df1.astype(int)
df2 = df2.astype(int)
#if first columns are not indices
#df1 = df1.set_index('ID')
#df2 = df2.set_index('ID')
df = df1.join(df2.sub(df1).add_prefix('sum'))
print (df)
A B sumA sumB
ID
0 10 2.0 5 3.0
1 11 3.0 6 5.0
2 12 4.0 7 5.0
Or similar:
df = df1.join(df2.sub(df1), rsuffix='sum')
print (df)
A B Asum Bsum
ID
0 10 2.0 5 3.0
1 11 3.0 6 5.0
2 12 4.0 7 5.0
Detail:
print (df2.sub(df1))
A B
ID
0 5 3.0
1 6 5.0
2 7 5.0
IIUC
df1[['C','D']]=(df2-df1)[['A','B']]
df1
Out[868]:
ID A B C D
0 0 10 2.0 5 3.0
1 1 11 3.0 6 5.0
2 2 12 4.0 7 5.0
df1.assign(B=0)
Out[869]:
ID A B C D
0 0 10 0 5 3.0
1 1 11 0 6 5.0
2 2 12 0 7 5.0
The 'ID' column should really be an index. See the Pandas tutorial on indexing for why this is a good idea.
df1 = df1.set_index('ID')
df2 = df2.set_index('ID')
df = df1.copy()
df[['C', 'D']] = df2 - df1
df['B'] = 0
print(df)
outputs
A B C D
ID
0 10 0 5 3.0
1 11 0 6 5.0
2 12 0 7 5.0