I have two data frames. One dataframe (A) looks like:
Name. gender start_coordinate end_coordinate ID
Peter M 30 150 1
Hugo M 4500 6000 2
Jennie F 300 700 3
The other dataframe (B) looks like
ID_sim. position string
1 89 aa
4 568 bb
5 938437 cc
I want to accomplish two tasks here:
I want to get a list of indices for rows (from dataframe B) for which position column falls in the interval (specified by start_coordinate and end_coordinate column) in dataframe A.
The result for this task will be:
lst = [0,1]. ### because row 0 of B falls in interval of row 1 in A and row 1 of B falls in interval of row 3 of A.
The indices that I get from task 1, I want to keep it from dataframe B to create a new dataframe. Thus, the new dataframe will look like:
position string
89 aa
568 bb
I used .between() to accomplish this task. The code is as follows:
lst=dfB[dfB['position'].between(dfA.loc[0,'start_coordinate'],dfA.loc[len(dfA)-1,'end_coordinate'])].index.tolist()
result=dfB[dfB.index.isin(lst)]
result.shape
However, when I run this piece of code I get the following error:
KeyError: 0
What could possibly be raising this error? And how can I solve this?
We can try numpy broadcasting here
s, e = dfA[['start_coordinate', 'end_coordinate']].to_numpy().T
p = dfB['position'].to_numpy()[:, None]
dfB[((p >= s) & (p <= e)).any(1)]
ID_sim. position string
0 1 89 aa
1 4 568 bb
You could use Pandas IntervalIndex to get the positions, and afterwards, use a boolean to pull the relevant rows from B :
Create IntervalIndex:
intervals = pd.IntervalIndex.from_tuples([*zip(A['start_coordinate'],
A['end_coordinate'])
],
closed='both')
Get indexers for B.position, create a boolean array with the values and filter B:
# get_indexer returns -1 if an index is not found.
B.loc[intervals.get_indexer(B.position) >= 0]
Out[140]:
ID_sim. position string
0 1 89 aa
1 4 568 bb
This should work. Less elegant but easier to comprehend.
import pandas as pd
data = [['Name.','gender', 'start_coordinate','end_coordinate','ID'],
['Peter','M',30,150,1],
['Hugo','M',4500,6000,2],
['Jennie','F',300,700,3]]
data2 = [['ID_sim.','position','string'],
['1',89,'aa'],
['4',568,'bb'],
['5',938437,'cc']]
df1 = pd.DataFrame(data[1:], columns=data[0])
df2 = pd.DataFrame(data2[1:], columns=data2[0])
merged = pd.merge(df1, df2, left_index=True, right_index=True)
print (merged[(merged['position'] > merged['start_coordinate']) & (merged['position'] < merged['end_coordinate'])])
Related
I have two data frames. One dataframe (dfA) looks like:
Name gender start_coordinate end_coordinate ID
Peter M 30 150 1
Hugo M 4500 6000 2
Jennie F 300 700 3
The other dataframe (dfB) looks like
Name position string
Peter 89 aa
Jennie 568 bb
Jennie 90 cc
I want to filter data from dfA such that position from dfB falls in the interval of dfA (start coordinate and end coordinate) and names should be same as well. For example, position value of row # 1 of dfB falls in interval specified by row # 1 of dfA and the corresponding name value is also the same therefore, I want this row. In contrast, row # 3 of dfB also falls in interval of row # 1 of dfA but the name value is different therefore, I don't want this record.
The expected out therefore, becomes:
##new_dfA
Name gender start_coordinate end_coordinate ID
Peter M 30 150 1
Jennie F 300 700 3
##new_dfB
Name position string
Peter 89 aa
Jennie 568 bb
In reality, dfB is of size (443068765,10) and dfA is of size (100000,3) therefore, I don't want to use numpy broadcasting because I run into memory error. Is there a way to deal with this problem within pandas framework. Insights will be appreciated.
If you have that many rows, pandas might not be well suited for your application.
That said, if there aren't many rows with identical "Name", you could merge on "Name" and then filter the rows matching your condition:
dfC = dfA.merge(dfB, on='Name')
dfC = dfC[dfC['position'].between(dfC['start_coordinate'], dfC['end_coordinate'])]
dfA_new = dfC[df1.columns]
dfB_new = dfC[df2.columns]
output:
>>> dfA_new
Name gender start_coordinate end_coordinate ID
0 Peter M 30 150 1
1 Jennie F 300 700 3
>>> dfB_new
Name position string
0 Peter 89 aa
1 Jennie 568 bb
use pandasql
pd.sql("select df1.* from df1 inner join df2 on df2.name=df1.name and df2.position between df1.start_coordinate and df1.end_coordinate",globals())
Name gender start_coordinate end_coordinate ID
0 Peter M 30 150 1
1 Jennie F 300 700 3
pd.sql("select df2.* from df1 inner join df2 on df2.name=df1.name and df2.position between df1.start_coordinate and df1.end_coordinate",globals())
Name position string
0 Peter 89 aa
1 Jennie 568 bb
I have a dataframe with 2M rows which is in the below format:
ID Number
1 30
1 40
1 60
2 10
2 30
3 60
I need to select the IDs have the number 30 and 40 present (in this case, output should be 1).
I know we can create a new DF having only numbers 30 & 40 and then groupby to see which IDs have more than count 1. But is there a way we can to do both in the groupby statement ?
My code:
a=df[(df['Number']==30) | (df['Number']==40) ]
b=a.groupby('ID')['Number'].nunique().to_frame(name='tt').reset_index()
b[b['tt'] > 1]
Use groupby filter and issubset
s = {30, 40}
df.groupby('ID').filter(lambda x: s.issubset(set(x.Number)))
Out[158]:
ID Number
0 1 30
1 1 40
2 1 60
I find the fact that the describe() method of Groupby objects returns a dataframe to be extremely helpful.
Output temp1 = a.groupby("ID").describe() and temp2 = a.groupby("ID").describe()["Number"] to a Jupyter notebook to see what they look like, then the following code (which follows on from yours) should make sense.
summary = a.groupby("ID").describe()["Number"]
summary.loc[summary["count"] > 1].index
I would create a df for each condition and then inner join them:
df1 = df[df.Number == 30][['Number']]
df2 = df[df.Number == 40][['Number']]
df3 = df1.join(df2,how='inner',on='Number')
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 am trying to update one dataframe with another dataframe with respect to the first column. If there is an extra row in the second dataframe, it should be inserted in the first dataframe. It there is a row with the same data in the first column but different data in the other coulmns, that row should be updated. Also, the row which has no value in the first column should be dropped.
Code used -
df = df_1.combine_first(df_2)\
.reset_index()\
.reindex(columns=df_1.columns)
df = df.drop_duplicates(subset='A', keep= 'last', inplace=False)
df.dropna(subset=['A'])
print ("Final Data")
print (df)
First Dataframe -
A B C
0 45 a b
1 98 c d
2 67 bn k
Second Dataframe -
A B C
0 45 a d
1 98 c d
2 67 bn k
3 90 x z
4
Final should look like -
A B C
0 45 a d
1 98 c d
2 67 bn k
3 90 x z
The final dataframe that I get -
A B C
0 45.0 a b
1 98.0 c d
2 67.0 bn k
3 90.0 x z
4
So, neither the data is getting updated, nor is it removing the row with null values. What am I missing?
Based on my understanding of your question, your second dataframe basically supercedes the first, if there is a matching index. If there isn't, then the difference is added to the first dataframe. I am also assuming that there are no duplicate keys in the first column, A.
Framing this requirement a little differently, the final output should contain all the rows in the second dataframe, as well as the values (since they are meant to overwrite the first dataframe if there's a match). Therefore, we will start off using the second dataframe as it is, and then add back the rows that exist in the first dataframe but not in the second. See the example below. (I'm also using a slightly different first dataframe to highlight the effects)
import pandas as pd
df1 = pd.DataFrame({'A':[45,98,67,91],'B':['a','c','bn','y'],'C':['b','d','k','oo']})
df2 = pd.DataFrame({'A':[45,98,67,90,''],'B':['a','c','bn','x',''],'C':['d','d','k','z','']})
# Remove rows with empty values in first column. This should be whatever conditions applicable to you i.e. checking for np.nan instead of str('')
df2 = df2.loc[df2['A'] != '']
df1.set_index('A', inplace=True)
df2.set_index('A', inplace=True)
# Find keys in dataframe 1 that are not in dataframe 2
idx_diff = df1.index.difference(df2.index)
# Append these rows to dataframe 2
df_ins = df1.loc[idx_diff]
df3 = df2.append(df_ins)
df3.reset_index(inplace=True)
>>>df3
A B C
0 45 a d
1 98 c d
2 67 bn k
3 90 x z
4 91 y oo
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')