I have a simple dataframe:
df = pd.DataFrame({'id': ['a','a','a','b','b'],'value':[0,15,20,30,0]})
df
id value
0 a 0
1 a 15
2 a 20
3 b 30
4 b 0
And I want a pivot table with the number of values greater than zero.
I tried this:
raw = pd.pivot_table(df, index='id',values='value',aggfunc=lambda x:len(x>0))
But returned this:
value
id
a 3
b 2
What I need:
value
id
a 2
b 1
I read lots of solutions with groupby and filter. Is it possible to achieve this only with pivot_table command? If it is not, which is the best approach?
Thanks in advance
UPDATE
Just to make it clearer why I am avoinding filter solution. In my real and complex df, I have other columns, like this:
df = pd.DataFrame({'id': ['a','a','a','b','b'],'value':[0,15,20,30,0],'other':[2,3,4,5,6]})
df
id other value
0 a 2 0
1 a 3 15
2 a 4 20
3 b 5 30
4 b 6 0
I need to sum the column 'other', but when i filter I got this:
df=df[df['value']>0]
raw = pd.pivot_table(df, index='id',values=['value','other'],aggfunc={'value':len,'other':sum})
other value
id
a 7 2
b 5 1
Instead of:
other value
id
a 9 2
b 11 1
Need sum for count Trues created by condition x>0:
raw = pd.pivot_table(df, index='id',values='value',aggfunc=lambda x:(x>0).sum())
print (raw)
value
id
a 2
b 1
As #Wen mentioned, another solution is:
df = df[df['value'] > 0]
raw = pd.pivot_table(df, index='id',values='value',aggfunc=len)
You can filter the dataframe before pivoting:
pd.pivot_table(df.loc[df['value']>0], index='id',values='value',aggfunc='count')
Related
i have 2 data frames where the index is the same value but different name. I need to add a column to DF1 but get that information from DF2
DF1:
SKU X Y Z
1234 0 0 0
5642 0 0 0
DF2:
AH SKU X Y Z Total
1234 4 5 1 10
5642 1 0 1 2
So I know I can add the total column to DF1 by doing
df1 ["Total"]
How can I now have that total from DF2 to DF1 making sure the SKU and AH SKU are matching?
You would do it like this, using default index matching.
df1["Total"] = df2["Total"]
That the indexes have different names shouldn't matter in this case (there's only one index level).
Make sure that those "columns" you pointed to in the answer really are the index: print(df1.index) should have something named SKU, and so on.
I am facing a situation where I need to group-by a dataframe by a column 'ID' and also calculate the total time frame depicted for that particular ID to complete. I only want to calculate the difference between the date_open and data_closed for the particular ID with the ID count.
We only need to focus on the date open and the date closed field. So it needs to do something taking the max closing date and the min open date and subtracting the two
The dataframe looks as follows:
ID Date_Open Date_Closed
1 01/01/2019 02/01/2019
1 07/01/2019 09/01/2019
2 10/01/2019 11/01/2019
2 13/01/2019 19/01/2019
3 10/01/2019 11/01/2019
The output should look like this :
ID Count_of_ID Total_Time_In_Days
1 2 8
2 2 9
3 1 1
How should I achieve this ?
Using GroupBy with named_aggregation and the min and max of the dates:
df[['Date_Open', 'Date_Closed']] = (
df[['Date_Open', 'Date_Closed']].apply(lambda x: pd.to_datetime(x, format='%d/%m/%Y'))
)
dfg = df.groupby('ID').agg(
Count_of_ID=('ID','size'),
Date_Open=('Date_Open','min'),
Date_Closed=('Date_Closed','max')
)
dfg['Total_Time_In_Days'] = dfg['Date_Closed'].sub(dfg['Date_Open']).dt.days
dfg = dfg.drop(columns=['Date_Closed', 'Date_Open']).reset_index()
ID Count_of_ID Total_Time_In_Days
0 1 2 8
1 2 2 9
2 3 1 1
Now we have Total_Time_In_Days as int:
print(dfg.dtypes)
ID int64
Count_of_ID int64
Total_Time_In_Days int64
dtype: object
This can also be used:
df['Date_Open'] = pd.to_datetime(df['Date_Open'], dayfirst=True)
df['Date_Closed'] = pd.to_datetime(df['Date_Closed'], dayfirst=True)
df_grouped = df.groupby(by='ID').count()
df_grouped['Total_Time_In_Days'] = df.groupby(by='ID')['Date_Closed'].max() - df.groupby(by='ID')['Date_Open'].min()
df_grouped = df_grouped.drop(columns=['Date_Open'])
df_grouped.columns=['Count', 'Total_Time_In_Days']
print(df_grouped)
Count Total_Time_In_Days
ID
1 2 8 days
2 2 9 days
3 1 1 days
I'll try first to create the a column depicting how much time passed from Date_open to Date_closed for each instance of the dataframe. Like this:
df['Total_Time_In_Days'] = df.Date_closed - df.Date_open
Then you can use groupby:
df.groupby('id').agg({'id':'count','Total_Time_In_Days':'sum'})
If you need any help with the .agg function you can refer to it's official documentation here.
How can I drop the whole group by city and district if date's value of 2018/11/1 not exits in the following dataframe:
city district date value
0 a c 2018/9/1 12
1 a c 2018/10/1 4
2 a c 2018/11/1 5
3 b d 2018/9/1 3
4 b d 2018/10/1 7
The expected result will like this:
city district date value
0 a c 2018/9/1 12
1 a c 2018/10/1 4
2 a c 2018/11/1 5
Thank you!
Create helper column by DataFrame.assign, compare by datetime and test if at least one true per groups with GroupBy.any and GroupBy.transform for possible filter by boolean indexing:
mask = (df.assign(new=df['date'].eq('2018/11/1'))
.groupby(['city','district'])['new'].transform('any'))
df = df[mask]
print (df)
city district date value
0 a c 2018/9/1 12
1 a c 2018/10/1 4
2 a c 2018/11/1 5
If error with misisng values in mask one possivle idea is replace misisng values in columns used for groups:
mask = (df.assign(new=df['date'].eq('2018/11/1'),
city= df['city'].fillna(-1),
district= df['district'].fillna(-1))
.groupby(['city','district'])['new'].transform('any'))
df = df[mask]
print (df)
city district date value
0 a c 2018/9/1 12
1 a c 2018/10/1 4
2 a c 2018/11/1 5
Another idea is add possible misisng index values by reindex and also replace missing values to False:
mask = (df.assign(new=df['date'].eq('2018/11/1'))
.groupby(['city','district'])['new'].transform('any'))
df = df[mask.reindex(df.index, fill_value=False).fillna(False)]
print (df)
city district date value
0 a c 2018/9/1 12
1 a c 2018/10/1 4
2 a c 2018/11/1 5
There's a special GroupBy.filter() method for this. Assuming date is already datetime:
filter_date = pd.Timestamp('2018-11-01').date()
df = df.groupby(['city', 'district']).filter(lambda x: (x['date'].dt.date == filter_date).any())
i have a dataframe in which I need to find a specific image name in the entire dataframe and sum its index values every time they are found. SO my data frame looks like:
c 1 2 3 4
g
0 180731-1-61.jpg 180731-1-61.jpg 180731-1-61.jpg 180731-1-61.jpg
1 1209270004-2.jpg 180609-2-31.jpg 1209270004-2.jpg 1209270004-2.jpg
2 1209270004-1.jpg 180414-2-38.jpg 180707-1-31.jpg 1209050002-1.jpg
3 1708260004-1.jpg 1209270004-2.jpg 180609-2-31.jpg 1209270004-1.jpg
4 1108220001-5.jpg 1209270004-1.jpg 1108220001-5.jpg 1108220001-2.jpg
I need to find the 1209270004-2.jpg in entire dataframe. And as it is found at index 1 and 3 I want to add the index values so it should be
1+3+1+1=6.
I tried the code:
img_fname = '1209270004-2.jpg'
df2 = df1[df1.eq(img_fname).any(1)]
sum = int(np.sum(df2.index.values))
print(sum)
I am getting the answer of sum 4 i.e 1+3=4. But it should be 6.
If the string occurence is only once or twice or thrice or four times like for eg 180707-1-31 is in column 3. then the sum should be 45+45+3+45 = 138. Which signifies that if the string is not present in the dataframe take vallue as 45 instead the index value.
You can multiple boolean mask by index values and then sum:
img_fname = '1209270004-1.jpg'
s = df1.eq(img_fname).mul(df1.index.to_series(), 0).sum()
print (s)
1 2
2 4
3 0
4 3
dtype: int64
out = np.where(s == 0, 45, s).sum()
print (out)
54
If dataset does not have many columns, this can also work with your original question
df1 = pd.DataFrame({"A":["aa","ab", "cd", "ab", "aa"], "B":["ab","ab", "ab", "aa", "ab"]})
s = 0
for i in df1.columns:
s= s+ sum(df1.index[df1.loc[:,i] == "ab"].tolist())
Input :
A B
0 aa ab
1 ab ab
2 cd ab
3 ab aa
4 aa ab
Output :11
Based on second requirement:
I have a dataframe like this:
In [24]: df = pd.DataFrame({'id': ['a','a','b','b','c','c'],'date':[201708,201709,201708,201709,201708,201709],'value':[0,15,20,30,20,0]})
In [25]: df
Out[25]:
date id value
0 201708 a 0
1 201709 a 15
2 201708 b 20
3 201709 b 30
4 201708 c 20
5 201709 c 0
And I have this derived pivot table:
In [26]: base=pd.pivot_table(df,index='id',columns='date',values='value',aggfunc='sum',fill_value=0,margins=False)
In [27]: base
Out[27]:
date 201708 201709
id
a 0 15
b 20 30
c 20 0
I need to create another df from this pivot table. In this new dataframe I need to show the values, for each id, that are larger than zero on date=t and evaluated as zero on the prior date(date=t-1). The result that I need is this df:
date 201708 201709
id
a 0 15
b 0 0
c 0 0
Does anyone know how to achieve this?
Thanks in advance.
Assuming your dataframe is df, use pd.DataFrame.where
df.where(
df.gt(0) & df.shift(axis=1).eq(0),
0
)