I am fairly new to Python and I am trying to create a new function to work on my project.
The function will aim to detect which unique value is present in another column of another table.
At first, the function seeks to keep only the unique values of the two tables, then merges them into a new dataframe
It's the rest that gets complicated because I would like to return which row and on which table my value is missing
If you have any other leads or thought patterns, I'm also interested.
Here is my code :
def correspondance_cle(df1, df2, col):
df11 = pd.DataFrame(df1[col].unique())
df11.columns= [col]
df11['test1'] = 1
df21 = pd.DataFrame(df2[col].unique())
df21.columns= [col]
df21['test2'] = 1
df3 = pd.merge(df11, df21, on=col, how='outer')
df3 = df3.loc[(fk['test1'].isna() == True) | (fk['test2'].isna() == True),:]
df3.info()
for row in df3[col]:
if df3['test1'].isna() == True:
print(row, "is not in df1")
else:
print(row, 'is not in df2')
Thanks to everyone who took the time to read the post.
First use outer join with remove duplicates by Series.drop_duplicates and Series.reset_index for avoid removed original indices:
df1 = pd.DataFrame({'a':[1,2,5,5]})
df2 = pd.DataFrame({'a':[2,20,5,8]})
col = 'a'
df = (df1[col].drop_duplicates().reset_index()
.merge(df2[col].drop_duplicates().reset_index(),
indicator=True,
how='outer',
on=col))
print (df)
index_x a index_y _merge
0 0.0 1 NaN left_only
1 1.0 2 0.0 both
2 2.0 5 2.0 both
3 NaN 20 1.0 right_only
4 NaN 8 3.0 right_only
Then filter rows by helper column _merge:
print (df[df['_merge'].eq('left_only')])
index_x a index_y _merge
0 0.0 1 NaN left_only
print (df[df['_merge'].eq('right_only')])
index_x a index_y _merge
3 NaN 20 1.0 right_only
4 NaN 8 3.0 right_only
Related
I have a dataframe as
col 1 col 2
A 2020-07-13
A 2020-07-15
A 2020-07-18
A 2020-07-19
B 2020-07-13
B 2020-07-19
C 2020-07-13
C 2020-07-18
I want it to become the following in a new dataframe
col_3 diff_btw_1st_2nd_date diff_btw_2nd_3rd_date diff_btw_3rd_4th_date
A 2 3 1
B 6 NaN NaN
C 5 NaN NaN
I tried getting the groupby at Col 1 level , but not getting the intended result. Can anyone help?
Use GroupBy.cumcount for counter pre column col 1 and reshape by DataFrame.set_index with Series.unstack, then use DataFrame.diff, remove first only NaNs columns by DataFrame.iloc, convert timedeltas to days by Series.dt.days per all columns and change columns names by DataFrame.add_prefix:
df['col 2'] = pd.to_datetime(df['col 2'])
df = (df.set_index(['col 1',df.groupby('col 1').cumcount()])['col 2']
.unstack()
.diff(axis=1)
.iloc[:, 1:]
.apply(lambda x: x.dt.days)
.add_prefix('diff_')
.reset_index())
print (df)
col 1 diff_1 diff_2 diff_3
0 A 2 3.0 1.0
1 B 6 NaN NaN
2 C 5 NaN NaN
Or use DataFrameGroupBy.diff with counter for new columns by DataFrame.assign, reshape by DataFrame.pivot and remove NaNs by c2 with DataFrame.dropna:
df['col 2'] = pd.to_datetime(df['col 2'])
df = (df.assign(g = df.groupby('col 1').cumcount(),
c1 = df.groupby('col 1')['col 2'].diff().dt.days)
.dropna(subset=['c1'])
.pivot('col 1','g','c1')
.add_prefix('diff_')
.rename_axis(None, axis=1)
.reset_index())
print (df)
col 1 diff_1 diff_2 diff_3
0 A 2.0 3.0 1.0
1 B 6.0 NaN NaN
2 C 5.0 NaN NaN
You can assign a cumcount number grouped by col 1, and pivot the table using that cumcount number.
Solution
df["col 2"] = pd.to_datetime(df["col 2"])
# 1. compute date difference in days using diff() and dt accessor
df["diff"] = df.groupby(["col 1"])["col 2"].diff().dt.days
# 2. assign cumcount for pivoting
df["cumcount"] = df.groupby("col 1").cumcount()
# 3. partial transpose, discarding the first difference in nan
df2 = df[["col 1", "diff", "cumcount"]]\
.pivot(index="col 1", columns="cumcount")\
.drop(columns=[("diff", 0)])
Result
# replace column names for readability
df2.columns = [f"d{i+2}-d{i+1}" for i in range(len(df2.columns))]
print(df2)
d2-d1 d3-d2 d4-d3
col 1
A 2.0 3.0 1.0
B 6.0 NaN NaN
C 5.0 NaN NaN
df after assing cumcount is like this
print(df)
col 1 col 2 diff cumcount
0 A 2020-07-13 NaN 0
1 A 2020-07-15 2.0 1
2 A 2020-07-18 3.0 2
3 A 2020-07-19 1.0 3
4 B 2020-07-13 NaN 0
5 B 2020-07-19 6.0 1
6 C 2020-07-13 NaN 0
7 C 2020-07-18 5.0 1
My table has 4 columns: order_id, item_id_1, item_id_2 and item_id_3. The three last columns cover the same type of information (the ids of products). I want to transform this table to get 2-columns table with "order_id" and "item_id", so my columns cover unique type of informations. That means, if in a particular order_id there were 3 products ordered, I will get three (instead of one) rows in my new table).
This will alow me, for exapmle, perform 'grupby' operation on 'item_id" column to count how meny times a particular product was ordered.
How this table transformation process is called?
For example, if you have a dataframe like this -
df = pd.DataFrame({'order_id':[1,2,3], 'item_id_1':['a','b','c'], 'item_id_2':['x','y',''], 'item_id_3':['','q','']})
df
order_id item_id_1 item_id_2 item_id_3
0 1 a x
1 2 b y q
2 3 c
pd.melt(df, id_vars=['order_id'], \
value_vars=['item_id_1', 'item_id_2', 'item_id_3'], \
var_name='item_id', value_name='item_value').\
replace('',np.nan).dropna().\
sort_values(['order_id']).\
reset_index(drop=True)\
[['order_id', 'item_id']]
So I'm not aware of any method that allows you to expand rows automatically as you're suggesting, but you can easily reach you're goal without. Let's start from a similar data frame, I put nan in cells of items that have not been ordered:
import pandas as pd
import numpy as np
data = {'order_id':[1,2,3],'item_id_1':[11,12,13],'item_id_2':[21,np.nan,23],'item_id_3':[31,np.nan,np.nan]}
df = pd.DataFrame(data)
cols = ['item_id_1','item_id_2','item_id_3']
print(df)
Out:
order_id item_id_1 item_id_2 item_id_3
0 1 11 21.0 31.0
1 2 12 NaN NaN
2 3 13 23.0 NaN
Then you can define a new empty data frame to fill by iterating through the rows of the initial one. For every item a new row is added to the empty data frame with same order_id and different item_id.
new_df = pd.DataFrame(columns = ['order_id','item_id']) # ,'item_num']
for ind, row in df.iterrows():
new_row = {}
new_row['order_id'] = row['order_id']
for col in cols: # for num, col in enumerate(cols):
item = row[col]
if not pd.isna(item):
new_row['item_id'] = item
# new_row['item_num'] = num +1
new_df = new_df.append(new_row,ignore_index=True)
print(new_df)
Out: # shape (6,2), ok because because 6 items have been ordered
order_id item_id
0 1.0 11.0
1 1.0 21.0
2 1.0 31.0
3 2.0 12.0
4 3.0 13.0
5 3.0 23.0
If you want, you could also add a third column to keep trace of the category of each item (i.e. if it was item_1, 2 or 3) by uncommenting the lines in the code, which gives you this output:
order_id item_id item_num
0 1.0 11.0 1.0
1 1.0 21.0 2.0
2 1.0 31.0 3.0
3 2.0 12.0 1.0
4 3.0 13.0 1.0
5 3.0 23.0 2.0
I am trying to write a function that will check a specified column for nulls, within a group in a dataframe. The example dataframe has two columns, ID and VALUE. Multiple rows exist per ID. I want to know if ANY of the rows for a particular ID have a NULL value in VALUE.
I have tried building the function with iterrows().
df = pd.DataFrame({'ID':[1,2,2,3,3,3],
'VALUE':[50,None,30,20,10,None]})
def nullValue(col):
for i, row in col.iterrows():
if ['VALUE'] is None:
return 1
else:
return 0
df2 = df.groupby('ID').apply(nullVALUE)
df2.columns = ['ID','VALUE','isNULL']
df2
I am expecting to retrieve a dataframe with three columns, ID, VALUE, and isNULL. If any row in a grouped ID has a null, all of the rows for that ID should have a 1 under isNull.
Example:
ID VALUE isNULL
1 50.0 0
2 NaN 1
2 30.0 1
3 20.0 1
3 10.0 1
3 NaN 1
A quick solution, borrowed partially from this answer is to use groupby with transform:
df = pd.DataFrame({'ID':[1,2,2,3,3,3,3],
'VALUE':[50,None,None,30,20,10,None]})
df['isNULL'] = (df.VALUE.isnull().groupby([df['ID']]).transform('sum') > 0).astype(int)
Out[51]:
ID VALUE isNULL
0 1 50.0 0
1 2 NaN 1
2 2 NaN 1
3 3 30.0 1
4 3 20.0 1
5 3 10.0 1
6 3 NaN 1
I need to delete the row completely in a dataframe having "None" value in all the columns. I am using the following code -
df.dropna(axis=0,how='all',thresh=None,subset=None,inplace=True)
This does not bring any difference to the dataframe. The rows with "None" value are still there.
How to achieve this?
There Nones should be strings, so use replace first:
df = df.replace('None', np.nan).dropna(how='all')
df = pd.DataFrame({
'a':['None','a', 'None'],
'b':['None','g', 'None'],
'c':['None','v', 'b'],
})
print (df)
a b c
0 None None None
1 a g v
2 None None b
df1 = df.replace('None', np.nan).dropna(how='all')
print (df1)
a b c
1 a g v
2 NaN NaN b
Or test values None with not equal and DataFrame.any:
df1 = df[df.ne('None').any(axis=1)]
print (df1)
a b c
1 a g v
2 None None b
You should be dropping in the axis 1. Use the how keyword to drop columns with any or all NaN values. Check the docs
import pandas as pd
import numpy as np
df = pd.DataFrame({'a':[1,2,3], 'b':[-1, 0, np.nan], 'c':[np.nan, np.nan, np.nan]})
df
a b c
0 1 -1.0 NaN
1 2 0.0 NaN
2 3 NaN 5.0
df.dropna(axis=1, how='any')
a
0 1
1 2
2 3
df.dropna(axis=1, how='all')
a b
0 1 -1.0
1 2 0.0
2 3 NaN
I was wondering how I can remove rows which have a negative value but keep the NaNs. At the moment I am using:
DF = DF.ix[DF['RAF01Time'] >= 0]
But this removes the NaNs.
Thanks in advance.
You need boolean indexing with another condition with isnull:
DF = DF[(DF['RAF01Time'] >= 0) | (DF['RAF01Time'].isnull())]
Sample:
DF = pd.DataFrame({'RAF01Time':[-1,2,3,np.nan]})
print (DF)
RAF01Time
0 -1.0
1 2.0
2 3.0
3 NaN
DF = DF[(DF['RAF01Time'] >= 0) | (DF['RAF01Time'].isnull())]
print (DF)
RAF01Time
1 2.0
2 3.0
3 NaN
Another solution with query:
DF = DF.query("~(RAF01Time < 0)")
print (DF)
RAF01Time
1 2.0
2 3.0
3 NaN
You can just use < 0 and then take the inverse of the condition.
DF = DF[~(DF['RAF01Time'] < 0)]