I have two columns in a dataframe. I want to create third column such that if first column > second column than 1 ow 0. As below
Df
Value1 value 2. Newcolumn
101. 0
97. 1
Comparing two columns in a Pandas DataFrame and write the results of the comparison to a third column. It can do easily by these syntaxes
conditions=[(condition1),(condition2)]
choices=["choice1","choice2"]
df["new_column_name"]=np.select(conditions, choices, default)
conditions are the conditions to check for between the two columns
choices are the results to return based on the conditions
np.select is used to return the results to the new column
The dataframe is:
import numpy as np
import pandas as pd
#create DataFrame
df = pd.DataFrame({'Value1': [100,101],
'value 2': [101,97]})
#define conditions
conditions = [df['Value1'] < df['value 2'],
df['Value1'] > df['value 2']]
#define choices
choices = ['0', '1']
#create new column in DataFrame that displays results of comparisons
df['Newcolumn'] = np.select(conditions, choices, default='Tie')
Final dataframe
print(df)
Output:
Value1 value 2 Newcolumn
0 100 101 0
1 101 97 1
Related
I have two dataframes
How would one populate the values in bold from df1 into the column 'Value' in df2?
Use melt on df1 before merge your 2 dataframes
tmp = df1.melt('Rating', var_name='Category', value_name='Value2')
df2['Value'] = df2.merge(tmp, on=['Rating', 'Category'])['Value2']
print(df2)
# Output
Category Rating Value
0 Hospitals A++ 2.5
1 Education AA 2.1
I have two dataframes. One is the basevales (df) and the other is an offset (df2).
How do I create a third dataframe that is the first dataframe offset by matching values (the ID) in the second dataframe?
This post doesn't seem to do the offset... Update only some values in a dataframe using another dataframe
import pandas as pd
# initialize list of lists
data = [['1092', 10.02], ['18723754', 15.76], ['28635', 147.87]]
df = pd.DataFrame(data, columns = ['ID', 'Price'])
offsets = [['1092', 100.00], ['28635', 1000.00], ['88273', 10.]]
df2 = pd.DataFrame(offsets, columns = ['ID', 'Offset'])
print (df)
print (df2)
>>> print (df)
ID Price
0 1092 10.02
1 18723754 15.76 # no offset to affect it
2 28635 147.87
>>> print (df2)
ID Offset
0 1092 100.00
1 28635 1000.00
2 88273 10.00 # < no match
This is want I want to produce: The price has been offset by matching
ID Price
0 1092 110.02
1 18723754 15.76
2 28635 1147.87
I've also looked at Pandas Merging 101
I don't want to add columns to the dataframe, and I don;t want to just replace column values with values from another dataframe.
What I want is to add (sum) column values from the other dataframe to this dataframe, where the IDs match.
The closest I come is df_add=df.reindex_like(df2) + df2 but the problem is that it sums all columns - even the ID column.
Try this :
df['Price'] = pd.merge(df, df2, on=["ID"], how="left")[['Price','Offset']].sum(axis=1)
My Dataframe contains 500 columns, but I only want to pick out 27 columns in a new Dataframe.
How do I do that?
I used query()
but output
TypeError: query() takes from 2 to 3 positional arguments but 27 were given
If you want to select the columns based on their name, you can do the following:
df_new = df[["colA", "colB", "colC", ...]]
or use the "filter" function:
df_new = df.filter(["colA", "colB", "colC", ..])
In case that your column selection is based on the index of columns:
df_new = df.iloc[:, 0:27] # if columns are consecutive
df_new = df.iloc[:, [0,2,10,..]] # if columns are not consecutive (the numbers refer to the column indices)
need one help.
I am trying to concatenate two data frames. 1st has 58k rows, other 100. Want to concatenate in a way that each of 58k row has 100 rows from other df. So in total 5.8 mil rows.
Performance is very poor, takes 1 hr to do 10 pct. Any suggestions for improvement?
Here is code snippet.
def myfunc(vendors3,cust_loc):
cust_loc_vend = pd.DataFrame()
cust_loc_vend.empty
for i,row in cust_loc.iterrows():
clear_output(wait=True)
a= row.to_frame().T
df= pd.concat([vendors3, a],axis=1, ignore_index=False)
#cust_loc_vend = pd.concat([cust_loc_vend, df],axis=1, ignore_index=False)
cust_loc_vend= cust_loc_vend.append(df)
print('Current progress:',np.round(i/len(cust_loc)*100,2),'%')
return cust_loc_vend
For e.g. if first DF has 5 rows and second has 100 rows
DF1 (sample 2 columns)
I want a merged DF such that each row in DF 2 has All rows from DF1-
Well all you are looking for is a join.But since there is no column column, what you can do is create a column which is similar in both the dataframes and then drop it eventually.
df['common'] = 1
df1['common'] = 1
df2 = pd.merge(df, df1, on=['common'],how='outer')
df = df.drop('tmp', axis=1)
where df and df1 are dataframes.
I tried following code to select columns from a dataframe. My dataframe has about 50 values. At the end, I want to create the sum of selected columns, create a new column with these sum values and then delete the selected columns.
I started with
columns_selected = ['A','B','C','D','E']
df = df[df.column.isin(columns_selected)]
but it said AttributeError: 'DataFrame' object has no attribute 'column'
Regarding the sum: As I don't want to write for the sum
df['sum_1'] = df['A']+df['B']+df['C']+df['D']+df['E']
I also thought that something like
df['sum_1'] = df[columns_selected].sum(axis=1)
would be more convenient.
You want df[columns_selected] to sub-select the df by a list of columns
you can then do df['sum_1'] = df[columns_selected].sum(axis=1)
To filter the df to just the cols of interest pass a list of the columns, df = df[columns_selected] note that it's a common error to just a list of strings: df = df['a','b','c'] which will raise a KeyError.
Note that you had a typo in your original attempt:
df = df.loc[:,df.columns.isin(columns_selected)]
The above would've worked, firstly you needed columns not column, secondly you can use the boolean mask as a mask against the columns by passing to loc or ix as the column selection arg:
In [49]:
df = pd.DataFrame(np.random.randn(5,5), columns=list('abcde'))
df
Out[49]:
a b c d e
0 -0.778207 0.480142 0.537778 -1.889803 -0.851594
1 2.095032 1.121238 1.076626 -0.476918 -0.282883
2 0.974032 0.595543 -0.628023 0.491030 0.171819
3 0.983545 -0.870126 1.100803 0.139678 0.919193
4 -1.854717 -2.151808 1.124028 0.581945 -0.412732
In [50]:
cols = ['a','b','c']
df.ix[:, df.columns.isin(cols)]
Out[50]:
a b c
0 -0.778207 0.480142 0.537778
1 2.095032 1.121238 1.076626
2 0.974032 0.595543 -0.628023
3 0.983545 -0.870126 1.100803
4 -1.854717 -2.151808 1.124028