I am using openpyxl to edit three dataframes, df1, df2, df3 (If it is necessary, we can also regard as three excels independently):
import pandas as pd
data1 = [[1, 1],[1,1]]
df1 = pd.DataFrame(data1, index = ['I1a','I1b'], columns=['v1a', 'v1b'])
df1.index.name='I1'
data2 = [[2, 2,2,2],[2,2,2,2],[2,2,2,2],[2,2,2,2]]
df2 = pd.DataFrame(data2, index = ['I2a','I2b','I2c','I2d'], columns=['v2a','v2b','v2c','v2d'])
df2.index.name='I2'
data3 = [['a', 'b',3,3],['a','c',3,3],['b','c',3,3],['c','d',3,3]]
df3 = pd.DataFrame(data3, columns=['v3a','v3b','v3c','v3d'])
df3 = df3.groupby(['v3a','v3b']).first()
Here df3 is multiindex. How to concat them into one excel horizontally (each dataframe start at the same line) as following:
Here we will regard index as a column and for multiindex, we keep the first level hidden.
Update
IIUC:
>>> pd.concat([df1.reset_index(), df2.reset_index(), df3.reset_index()], axis=1)
I1 v1a v1b I2 v2a v2b v2c v2d v3a v3b v3c v3d
0 I1a 1.0 1.0 I2a 2 2 2 2 a b 3 3
1 I1b 1.0 1.0 I2b 2 2 2 2 a c 3 3
2 NaN NaN NaN I2c 2 2 2 2 b c 3 3
3 NaN NaN NaN I2d 2 2 2 2 c d 3 3
Old answer
Assuming you know the start row, you can use pandas to remove extra columns:
import pandas as pd
df = pd.read_excel('input.xlsx', header=0, skiprows=3).dropna(how='all', axis=1)
df.to_excel('output.xlsx', index=False)
Input:
Output:
Related
I have two dataframes like df1, df2.
In df1 i have 4 columns (A,B,C,D) and two rows,
In df2 i have 4 columns (A,B,C,D) and two rows.
Now I want to subtract the two dataframe LIKE df1['A'] - df2['A'] and so on. But I don't know how to do it.
df1-
df2 -
Just do the subtraction but keep in mind the indexes, for example, let's say I have df1 and df2 with same columns but different index:
df1 = dd.from_array(np.arange(8).reshape(2, 4), columns=['A','B','C','D'])
df2 = dd.from_pandas(pd.DataFrame(
np.arange(8).reshape(2, 4),
columns=['A','B','C','D'],
index=[1, 2]
), npartitions=1)
Then:
(df1 - df2).compute()
# A B C D
# 0 NaN NaN NaN NaN
# 1 4.0 4.0 4.0 4.0
# 2 NaN NaN NaN NaN
On the other hand, let's match index from df2 to df1 and subtract
df2 = df2.assign(idx=1)
df2 = df2.set_index(df2.idx.cumsum() - 1)
df2 = df2.drop(columns=['idx'])
(df1 - df2).compute()
# A B C D
# 0 0 0 0 0
# 1 0 0 0 0
I have two dataframes containing the same columns:
df1 = pd.DataFrame({'a': [1,2,3,4,5],
'b': [2,3,4,5,6]})
df2 = pd.DataFrame({'a': [1,3,4],
'b': [2,4,5]})
I want df2 to have the same number of rows as df1. Any values of a not present in df1 should be copied over, and corresponding values of b should be taken from the row before.
In other words, I want df2 to look like this:
df2 = pd.DataFrame({'a': [1,2,3,4,5],
'b': [2,2,4,5,5]})
EDIT: I'm looking for an answer that is independent of the number of columns
Use DataFrame.merge by only a column from df1 and for replace missing values is added forward filling them:
df = df1[['a']].merge(df2, how='left', on='a').ffill()
print (df)
a b
0 1 2.0
1 2 2.0
2 3 4.0
3 4 5.0
4 5 5.0
Or use merge_asof:
df = pd.merge_asof(df1[['a']], df2, on='a')
print (df)
a b
0 1 2
1 2 2
2 3 4
3 4 5
4 5 5
I'm in a trouble with adding a new column to a pandas dataframe when the length of new column value is bigger than length of index.
Data may like this :
import pandas as pd
df = pd.DataFrame(
{
"bar": ["A","B","C"],
"zoo": [1,2,3],
})
So, you see, length of this df's index is 3.
And next I wanna add a new column , code may like this two ways below:
df["new_col"] = [1,2,3,4]
It'll raise an error : Length of values does not match length of index.
Or:
df["new_col"] = pd.Series([1,2,3,4])
I will just get values[1,2,3] in my data frame df.
(The count of new column values can't out of the max index).
Now, what I want just like :
Is there a better way ?
Looking forward to your answer,thanks!
Use DataFrame.join with change Series name and right join:
#if not default index
#df = df.reset_index(drop=True)
df = df.join(pd.Series([1,2,3,4]).rename('new_col'), how='right')
print (df)
bar zoo new_col
0 A 1.0 1
1 B 2.0 2
2 C 3.0 3
3 NaN NaN 4
Another idea is add reindex by new s.index:
s = pd.Series([1,2,3,4])
df = df.reindex(s.index)
df["new_col"] = s
print (df)
bar zoo new_col
0 A 1.0 1
1 B 2.0 2
2 C 3.0 3
3 NaN NaN 4
s = pd.Series([1,2,3,4])
df = df.reindex(s.index).assign(new_col = s)
df = pd.DataFrame(
{
"bar": ["A","B","C"],
"zoo": [1,2,3],
})
new_col = pd.Series([1,2,3,4])
df = pd.concat([df,new_col],axis=1)
print(df)
bar zoo 0
0 A 1.0 1
1 B 2.0 2
2 C 3.0 3
3 NaN NaN 4
Input DF:
df = pd.DataFrame({'A': ['one',np.nan,'two',np.nan],
'B': [np.nan,22,np.nan,44],
'group':[0,0,1,1]
})
print(df)
A B group
0 one NaN 0
1 NaN 22.0 0
2 two NaN 1
3 NaN 44.0 1
I want to merge those rows in one, all cells in one in same column. But taking into account groups.
Currently have:
df=df.agg(lambda x: ','.join(x.dropna().astype(str))
).to_frame().T
print(df)
A B group
0 one,two 22.0,44.0 0,0,1,1
but this way is taking all rows, not only groups
Expected Output:
A B
0 one 22.0
1 two 44.0
If possible simplify solution for first non missing value per group use:
df = df.groupby('group').first()
print(df)
A B
group
0 one 22.0
1 two 44.0
If not and need general solution:
df = pd.DataFrame({'A': ['one',np.nan,'two',np.nan],
'B': [np.nan,22,np.nan,44],
'group':[0,0,0,1]
})
def f(x):
return x.apply(lambda x: pd.Series(x.dropna().to_numpy()))
df = df.set_index('group').groupby('group').apply(f).reset_index(level=1, drop=True).reset_index()
print(df)
group A B
0 0 one 22.0
1 0 two NaN
2 1 NaN 44.0
df_a = df.drop('B', axis=1).dropna()
df_b = df.drop('A', axis=1).dropna()
pd.merge(df_a, df_b, on='group')
I want to append a Series to a DataFrame where Series's index matches DataFrame's columns using pd.concat, but it gives me surprises:
df = pd.DataFrame(columns=['a', 'b'])
sr = pd.Series(data=[1,2], index=['a', 'b'], name=1)
pd.concat([df, sr], axis=0)
Out[11]:
a b 0
a NaN NaN 1.0
b NaN NaN 2.0
What I expected is of course:
df.append(sr)
Out[14]:
a b
1 1 2
It really surprises me that pd.concat is not index-columns aware. So is it true that if I want to concat a Series as a new row to a DF, then I can only use df.append instead?
Need DataFrame from Series by to_frame and transpose:
a = pd.concat([df, sr.to_frame(1).T])
print (a)
a b
1 1 2
Detail:
print (sr.to_frame(1).T)
a b
1 1 2
Or use setting with enlargement:
df.loc[1] = sr
print (df)
a b
1 1 2
"df.loc[1] = sr" will drop the column if it isn't in df
df = pd.DataFrame(columns = ['a','b'])
sr = pd.Series({'a':1,'b':2,'c':3})
df.loc[1] = sr
df will be like:
a b
1 1 2