I am trying to reassign multiple columns in DataFrame with modifications.
The below is a simplified example.
import pandas as pd
d = {'col1':[1,2], 'col2':[3,4]}
df = pd.DataFrame(d)
print(df)
col1 col2
0 1 3
1 2 4
I use assign() method to add 1 to both 'col1' and 'col2'.
However, the result is to add 1 only to 'col2' and copy the result to 'col1'.
df2 = df.assign(**{c: lambda x: x[c] + 1 for c in ['col1','col2']})
print(df2)
col1 col2
0 4 4
1 5 5
Can someone explain why this is happening, and also suggest a correct way to apply assign() to multiple columns?
I think the lambda here can not be used within the for loop dict
df.assign(**{c: df[c] + 1 for c in ['col1','col2']})
Related
I am trying to find a better, more pythonic way of accomplishing the following:
I want to add a new column to business_df called 'dot_prod', which is the dot product of a fixed vector (fixed_vector) and a vector from another data frame (rating_df). The rows of both business_df and rating_df have the same index values (business_id).
I have this loop which appears to work, however I know it's super clumsy (and takes forever). Essentially it loops through once for every row, calculates the dot product, then dumps it into the business_df dataframe.
n=0
for i in range(business_df.shape[0]):
dot_prod = np.dot(fixed_vector, rating_df.iloc[n])
business_df['dot_prod'][n] = dot_prod
n+=1
IIUC, you are looking for apply across axis=1 like:
business_df['dot_prod'] = rating_df.apply(lambda x: np.dot(fixed_vector, x), axis=1)
>>> fixed_vector = [1, 2, 3]
>>> df = pd.DataFrame({'col1' : [1,2], 'col2' : [3,4], 'col3' : [5,6]})
>>> df
col1 col2 col3
0 1 3 5
1 2 4 6
>>> df['col4'] = np.dot(fixed_vector, [df['col1'], df['col2'], df['col3']])
>>> df
col1 col2 col3 col4
0 1 3 5 22
1 2 4 6 28
I have the following question and I need help to apply the for loop to iterate through dataframe columns with unique values. For ex I have the following df.
col1 col2 col3
aaa 10 1
bbb 15 2
aaa 12 1
bbb 16 3
ccc 20 3
ccc 50 1
ddd 18 2
I had to apply some manipulation to the dataset for each unique value of col3. Therefore, what I did is I sliced out the df with col3=1 by:
df1 = df[df['col3']==1]
#added all processing here in df1#
Now I need to do the same slicing for col3==2 ... col3==10, and I will be applying the same manipulation as I did in col3==1. For ex I have to do:
df2 = df[df['col3']==2]
#add the same processing here in df2#
df3 = df[df['col3']==3]
#add the same processing here in df3#
Then I will need to append them into a list and then combine them at the end.
I couldn't figure out how to run a for loop that will go through col3 column and look at the unique values so I don't have to create manually ten dfs.
I tried to groupby then apply the manipulation but it didn't work.
I appreciate help on this. Thanks
simple solution. just iterate on the unique values of this column and loc the rows with this unique value. like this:
dfs=[]
for i in df["col3"].unique():
df_i = df.loc[df["Cluster"]==i,:]
dfs.append(df_i.copy())
This should do it but will be slow for large dataframes.
df1 = pd.DataFrame(columns=['col1', 'col2', 'col3'])
df2 = pd.DataFrame(columns=['col1', 'col2', 'col3'])
df3 = pd.DataFrame(columns=['col1', 'col2', 'col3'])
for _, v in df.iterrows():
if v[2] == 1:
# add your code
df1 = df1.append(v)
elif v[2] == 2:
# add your code
df2 = df2.append(v)
elif v[2] == 3:
# add your code
df3 = df3.append(v)
You can then use pd.concat() to rebuild to one df.
Output of df1
col1 col2 col3
0 aaa 10 1
2 aaa 12 1
5 ccc 50 1
My question is about the pandas.DataFrame.filter command. It seems that pandas creates a copy of the data frame to write any changes. How am I able to write on the data frame itself?
In other words:
d = {'col1': [1, 2], 'col2': [3, 4]}
df = pd.DataFrame(data=d)
df.filter(regex='col1').iloc[0]=10
Output:
col1 col2
0 1 3
1 2 4
Desired Output:
col1 col2
0 10 3
1 2 4
I think you need extract columns names and then use loc or iloc functions:
cols = df.filter(regex='col1').columns
df.loc[0, cols]=10
Or:
df.iloc[0, df.columns.get_indexer(cols)] = 10
print (df)
col1 col2
0 10 3
1 2 4
You cannnot use filter function, because subset returns a Series/DataFrame which may have its data as a view. That's why SettingWithCopyWarning is possible there (or raise if you set the option).
suppose a dataframe like this one:
df = pd.DataFrame([[1,2,3,4],[5,6,7,8],[9,10,11,12]], columns = ['A', 'B', 'A1', 'B1'])
I would like to have a dataframe which looks like:
what does not work:
new_rows = int(df.shape[1]/2) * df.shape[0]
new_cols = 2
df.values.reshape(new_rows, new_cols, order='F')
of course I could loop over the data and make a new list of list but there must be a better way. Any ideas ?
The pd.wide_to_long function is built almost exactly for this situation, where you have many of the same variable prefixes that end in a different digit suffix. The only difference here is that your first set of variables don't have a suffix, so you will need to rename your columns first.
The only issue with pd.wide_to_long is that it must have an identification variable, i, unlike melt. reset_index is used to create a this uniquely identifying column, which is dropped later. I think this might get corrected in the future.
df1 = df.rename(columns={'A':'A1', 'B':'B1', 'A1':'A2', 'B1':'B2'}).reset_index()
pd.wide_to_long(df1, stubnames=['A', 'B'], i='index', j='id')\
.reset_index()[['A', 'B', 'id']]
A B id
0 1 2 1
1 5 6 1
2 9 10 1
3 3 4 2
4 7 8 2
5 11 12 2
You can use lreshape, for column id numpy.repeat:
a = [col for col in df.columns if 'A' in col]
b = [col for col in df.columns if 'B' in col]
df1 = pd.lreshape(df, {'A' : a, 'B' : b})
df1['id'] = np.repeat(np.arange(len(df.columns) // 2), len (df.index)) + 1
print (df1)
A B id
0 1 2 1
1 5 6 1
2 9 10 1
3 3 4 2
4 7 8 2
5 11 12 2
EDIT:
lreshape is currently undocumented, but it is possible it might be removed(with pd.wide_to_long too).
Possible solution is merging all 3 functions to one - maybe melt, but now it is not implementated. Maybe in some new version of pandas. Then my answer will be updated.
I solved this in 3 steps:
Make a new dataframe df2 holding only the data you want to be added to the initial dataframe df.
Delete the data from df that will be added below (and that was used to make df2.
Append df2 to df.
Like so:
# step 1: create new dataframe
df2 = df[['A1', 'B1']]
df2.columns = ['A', 'B']
# step 2: delete that data from original
df = df.drop(["A1", "B1"], 1)
# step 3: append
df = df.append(df2, ignore_index=True)
Note how when you do df.append() you need to specify ignore_index=True so the new columns get appended to the index rather than keep their old index.
Your end result should be your original dataframe with the data rearranged like you wanted:
In [16]: df
Out[16]:
A B
0 1 2
1 5 6
2 9 10
3 3 4
4 7 8
5 11 12
Use pd.concat() like so:
#Split into separate tables
df_1 = df[['A', 'B']]
df_2 = df[['A1', 'B1']]
df_2.columns = ['A', 'B'] # Make column names line up
# Add the ID column
df_1 = df_1.assign(id=1)
df_2 = df_2.assign(id=2)
# Concatenate
pd.concat([df_1, df_2])
I'm trying to wrap my head around Pandas groupby methods. I'd like to write a function that does some aggregation functions and then returns a Pandas DataFrame. Here's a grossly simplified example using sum(). I know there are easier ways to do simple sums, in real life my function is more complex:
import pandas as pd
df = pd.DataFrame({'col1': ['A', 'A', 'B', 'B'], 'col2':[1.0, 2, 3, 4]})
In [3]: df
Out[3]:
col1 col2
0 A 1
1 A 2
2 B 3
3 B 4
def func2(df):
dfout = pd.DataFrame({ 'col1' : df['col1'].unique() ,
'someData': sum(df['col2']) })
return dfout
t = df.groupby('col1').apply(func2)
In [6]: t
Out[6]:
col1 someData
col1
A 0 A 3
B 0 B 7
I did not expect to have col1 in there twice nor did I expect that mystery index looking thing. I really thought I would just get col1 & someData.
In my real life application I'm grouping by more than one column and really would like to get back a DataFrame and not a Series object.
Any ideas for a solution or an explanation on what Pandas is doing in my example above?
----- added info -----
I should have started with this example, I think:
In [13]: import pandas as pd
In [14]: df = pd.DataFrame({'col1':['A','A','A','B','B','B'], 'col2':['C','D','D','D','C','C'], 'col3':[.1,.2,.4,.6,.8,1]})
In [15]: df
Out[15]:
col1 col2 col3
0 A C 0.1
1 A D 0.2
2 A D 0.4
3 B D 0.6
4 B C 0.8
5 B C 1.0
In [16]: def func3(df):
....: dfout = sum(df['col3']**2)
....: return dfout
....:
In [17]: t = df.groupby(['col1', 'col2']).apply(func3)
In [18]: t
Out[18]:
col1 col2
A C 0.01
D 0.20
B C 1.64
D 0.36
In the above illustration the result of the apply() function is a Pandas Series. And it lacks the groupby columns from the df.groupby. The essence of what I'm struggling with is how do I create a function which I apply to a groupby which returns both the result of the function AND the columns on which it was grouped?
----- yet another update ------
It appears that if I then do this:
pd.DataFrame(t).reset_index()
I get back a dataframe which is really close to what I was after.
The reason you are seeing the columns with 0s is because the output of .unique() is an array.
The best way to understand how your apply is going to work is to inspect each action group-wise:
In [11] :g = df.groupby('col1')
In [12]: g.get_group('A')
Out[12]:
col1 col2
0 A 1
1 A 2
In [13]: g.get_group('A')['col1'].unique()
Out[13]: array([A], dtype=object)
In [14]: sum(g.get_group('A')['col2'])
Out[14]: 3.0
The majority of the time you want this to be an aggregated value.
The output of grouped.apply will always have the group labels as an index (the unique values of 'col1'), so your example construction of col1 seems a little obtuse to me.
Note: To pop 'col1' (the index) back to a column you can call reset_index, so in this case.
In [15]: g.sum().reset_index()
Out[15]:
col1 col2
0 A 3
1 B 7