I have a data frame(DF1) with 100 columns.( one of the column is ID)
I have one more data frame(DF2) with 30 columns.( one column is ID)
I have to update the first 30 columns of the data frame(DF1) with the values in second data frame (DF2) keeping the rest of the values in the remaining columns of first data frame (DF1) intact.
update the first 30 column value in DF1 out of the 100 columns when the ID in second data frame (DF2) is present in first data frame (DF1).
I tested this on Python 3.7 but I see no reason for it not to work on 2.7:
joined = df1.reset_index() \
[['index', 'ID']] \
.merge(df2, on='ID')
df1.loc[joined['index'], df1.columns[:30]] = joined.drop(columns=['index', 'ID'])
This assumes that df2 doesn't have a column called index or the merge will fail saying duplicate key with suffix.
Here a slow-motion of its inner workings:
df1.reset_index() returns a dataframe same as df1 but with an additional column: index
[['index', 'ID']] extracts a dataframe containing just these 2 columns from the dataframe in #1
.merge(...) merges with df2 , matching on ID . The result (joined) is a dataframe with 32 columns: index, ID and the original 30 columns of df2.
df1.loc[<row_indexes>, <column_names>] = <another_dataframe> mean you want to replace at those particular cells with data from another_dataframe. Since joined has 32 columns, we need to drop the extra 2 (index and ID)
Related
I have a dataframe with multiple columns as t_orno,t_pono, t_sqnb ,t_pric,....and so on(it's a table with multiple columns).
The 2nd dataframe contains certain name of the columns from 1st dataframe. Eg.
columnname
t_pono
t_pric
:
:
I need to select only those columns from the 1st dataframe whose name is present in the 2nd. In above example t_pono,t_pric.
How can this be done?
Let's say you have the following columns (which can be obtained using df.columns, which returns a list):
df1_cols = ["t_orno", "t_pono", "t_sqnb", "t_pric"]
df2_cols = ["columnname", "t_pono", "t_pric"]
To get only those columns from the first dataframe that are present in the second one, you can do set intersection (and I cast it to a list, so it can be used to select data):
list(set(df1_cols).intersection(df2_cols))
And we get the result:
["t_pono", "t_pric"]
To put it all together and select only those columns:
select_columns = list(set(df1_cols).intersection(df2_cols))
new_df = df1.select(*select_columns)
I have two dataframes.
df1 has an index list made of strings like (row1,row2,..,rown) and a column list made of strings like (col1,col2,..,colm) while df2 has k rows and 3 columns (char_1,char_2,value). char_1 contains strings like df1 indexes while char_2 contains strings like df1 columns. I only want to assign the df2 value to df1 in the right position. For example if the first row of df2 reads ['row3','col1','value2'] I want to assign value2 to df1 in the position ([2,0]) (third row and first column).
I tried to use two functions to slide rows and columns of df1:
def func1(val):
# first I convert the series to dataframe
val=val.to_frame()
val=val.reset_index()
val=val.set_index('index') # I set the index so that it's the right column
def func2(val2):
try: # maybe the combination doesn't exist
idx1=list(cou.index[df2[char_2]==(val2.name)]) #val2.name reads col name of df1
idx2=list(cou.index[df2[char_1]==val2.index.values[0]]) #val2.index.values[0] reads index name of df1
idx= list(reduce(set.intersection, map(set, [idx1,idx2])))
idx=int(idx[0]) # final index of df2 where I need to take value to assign to df1
check=1
except:
check=0
if check==1: # if index exists
val2[0]=df2['value'][idx] # assign value to df1
return val2
val=val.apply(func2,axis=1) #apply the function for columns
val=val.squeeze() #convert again to series
return val
df1=df1.apply(func1,axis=1) #apply the function for rows
I made the conversion inside func1 because without this step I wasn't able to work with series keeping index and column names so I wasn't able to find the index idx in func2.
Well the problem is that it takes forever. df1 size is (3'600 X 20'000) and df2 is ( 500 X 3 ) so it's not too much. I really don't understand the problem.. I run the code for the first row and column to check the result and it's fine and it takes 1 second, but now for the entire process I've been waiting for hours and it's still not finished.
Is there a way to optimize it? As I wrote in the title I only need to run a function that keeps column and index names and works sliding the entire dataframe. Thanks in advance!
One Df1 with all data, one of the column with vl1,vl3,vl4,vl2,vl3,vl4..etc and another dataframe valueDf with a single column which have unique values vl1,vl2,vl3..etc
I want to split valueDf into an array of 4 dataframes like
Df(0)=(vl1,vl4,vl5)
Df(1)=(vl3,vl2,vl6)
Df(2)=(vl7,vl8,vl9)... etc
So that i can do leftsemi join on Df1 and write in 4 different locations
/folder_Df(0)/
/foldwer_Df(1)/.. Etc
I tried doing a randomSplit as below
val Dfs = valueDf.randomSplit(Array.range(1,5).map(_.toDouble), 1)
But now I have set of values which need to go into Df(0),Df(1),Df(2) based on this list (vl1,vl4,vl5),(vl3,vl2,vl6),(vl7,vl8,vl9).
I have two data frames, based on a condition that I get from a list (which length is 2 million) I get rows that match that condition, then for those rows I change the values in columns x and y in the first data frame by the values of x and y in the second data frame. Here is my code, but it is very slow and makes my computer freeze. Any idea how I can do this more efficiently ?
for ids in List_id:
a=df1.index[(df1['id'] == ids )==True].values[0]
b=df2.index[(df2['id'] == ids )==True].values[0]
df1['x'][a] = df2['x'][b]
df1['y'][a] = df2['y'][b]
thank you
--
Example:
List_id=[1, 11 , 12 , 13]
ids=1
a=df1.index[(df1['id'] == 1 )==True].values[0]
print('a') : 234
b=df2.index[(df2['id'] == 1 )==True].values[0]
print('b') : 789
df1['x'][a] = 0
df2['x'][b] =15
So at the end I want in my data frame 1:
df1['x'][a] = df2['x'][b]
Assuming you don't have repeated id in both dataframe, you can try something like below:
step-1: filtering df2
step-2: joining df1 with filtered one
step-3 replace values in joined df and dropping extra columns.
df2_filtered=df2[df2['id'].isin(List_id)]
join_df = df1.setIndex('id').join(df2_filtered.setIndex('id'), rsuffix = "_ignore", how = 'left')
# other columns from df2 will be null, you can use that to get the rows which needs to be updated
I am selecting row by row as follows:
for i in range(num_rows):
row = df.iloc[i]
as a result I am getting a Series object where row.index.values contains names of df columns.
But I wanted instead dataframe with only one row having dataframe columns in place.
When I do row.to_frame() instead of 1x85 dataframe (1 row, 85 cols) I get 85x1 dataframe where index contains names of columns and row.columns
outputs
Int64Index([0], dtype='int64').
But all I want is just original data-frame columns with only one row. How do I do it?
Or how do I convert row.index values to row.column values and change 85x1 dimension to 1x85
You just need to adding T
row.to_frame().T
Also change your for loop with adding []
for i in range(num_rows):
row = df.iloc[[i]]