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Insert Row in Dataframe at certain place

I have the following Dataframe:
Now i want to insert an empty row after every time the column "Zweck" equals 7.
So for example the third row should be an empty row.

import numpy as np
import pandas as pd
df = pd.DataFrame({'a': [1, 2, 3, 4, 5], 'b': [1, 2, 3, 4, 5], 'f': [1, 7, 3, 4, 7]})
ren_dict = {i: df.columns[i] for i in range(len(df.columns))}
ind = df[df['f'] == 7].index
df = pd.DataFrame(np.insert(df.values, ind, values=[33], axis=0))
df.rename(columns=ren_dict, inplace=True)
ind_empt = df['a'] == 33
df[ind_empt] = ''
print(df)
Output
a b f
0 1 1 1
1
2 2 2 7
3 3 3 3
4 4 4 4
5
6 5 5 7
Here the dataframe is overwritten, as the append operation will be resource intensive. As a result, the required strings with values 33 appear. This is necessary because np.insert does not allow string values to be substituted. Columns are renamed to their original state with: df.rename. Finally, we find lines with df['a'] == 33 to set to empty values.

How to name Pandas Dataframe Columns automatically?

I have a Pandas dataframe df with 102 columns. Each column is named differently, say A, B, C etc. to give the original dataframe following structure
Column A. Column B. Column C. ....
Row 1.
Row 2.
---
Row n
I would like to change the columns names from A, B, C etc. to F1, F2, F3, ...., F102. I tried using df.columns but wasn't successful in renaming them this way. Any simple way to automatically rename all column names to F1 to F102 automatically, insteading of renaming each column name individually?

df.columns=["F"+str(i) for i in range(1, 103)]
Note:
Instead of a “magic” number 103 you may use the calculated number of columns (+ 1), e.g.
len(df.columns) + 1, or
df.shape[1] + 1.
(Thanks to ALollz for this tip in his comment.)

One way to do this is to convert it to a pair of lists, and convert the column names list to the index of a loop:
import pandas as pd
d = {'Column A': [1, 2, 3, 4, 5, 4, 3, 2, 1], 'Column B': [1, 2, 3, 4, 5, 4, 3, 2, 1], 'Column c': [1, 2, 3, 4, 5, 4, 3, 2, 1]}
dataFrame = pd.DataFrame(data=d)
cols = list(dataFrame.columns.values) #convert original dataframe into a list containing the values for column name
index = 1 #start at 1
for column in cols:
cols[index-1] = "F"+str(index) #rename the column name based on index
index += 1 #add one to index
vals = dataFrame.values.tolist() #get the values for the rows
newDataFrame = pd.DataFrame(vals, columns=cols) #create a new dataframe containing the new column names and values from rows
print(newDataFrame)
Output:
F1 F2 F3
0 1 1 1
1 2 2 2
2 3 3 3
3 4 4 4
4 5 5 5
5 4 4 4
6 3 3 3
7 2 2 2
8 1 1 1

Select rows based on condition and set values from a vector

I want to set the entire rows to a value from a vector, if a condition in on column is met.
import pandas as pd
df = pd.DataFrame([['a', 1, 1], ['a', 1, 1], ['b', 1, 1]], columns=('one', 'two', 'three'))
vector = pd.Series([2,3,4])
print(df)
one two three
0 a 1 1
1 a 1 1
2 b 1 1
I want the result to be like this:
df_wanted = pd.DataFrame([['a', 1, 1], ['a', 1, 1], ['b', 4, 4]], columns=('one', 'two', 'three'))
print(df_wanted)
one two three
0 a 1 1
1 a 1 1
2 b 4 4
I tried this but it gives me error:
df.loc[df['one']=='b'] = vector[df['one']=='b']
ValueError: Must have equal len keys and value when setting with an iterable
// m.

You can specify columns in list for set:
df.loc[df['one']=='b', ['two', 'three']] = vector[df['one']=='b']
print(df)
one two three
0 a 1 1
1 a 1 1
2 b 4 4
Or if need more dynamic solution - select all numeric columns:
df.loc[df['one']=='b', df.select_dtypes(np.number).columns] = vector[df['one']=='b']
Or compare only once and assign to variable:
m = df['one']=='b'
df.loc[m, df.select_dtypes(np.number).columns] = vector[m]

Creating Pivot DataFrame using Multiple Columns in Pandas

I have a pandas dataframe following the form in the example below:
data = {'id': [1,1,1,1,2,2,2,2,3,3,3], 'a': [-1,1,1,0,0,0,-1,1,-1,0,0], 'b': [1,0,0,-1,0,1,1,-1,-1,1,0]}
df = pd.DataFrame(data)
Now, what I want to do is create a pivot table such that for each of the columns except the id, I will have 3 new columns corresponding to the values. That is, for column a, I will create a_neg, a_zero and a_pos. Similarly, for b, I will create b_neg, b_zero and b_pos. The values for these new columns would correspond to the number of times those values appear in the original a and b column. The final dataframe should look like this:
result = {'id': [1,2,3], 'a_neg': [1, 1, 1],
'a_zero': [1, 2, 2], 'a_pos': [2, 1, 0],
'b_neg': [1, 1, 1], 'b_zero': [2,1,1], 'b_pos': [1,2,1]}
df_result = pd.DataFrame(result)
Now, to do this, I can do the following steps and arrive at my final answer:
by_a = df.groupby(['id', 'a']).count().reset_index().pivot('id', 'a', 'b').fillna(0).astype(int)
by_a.columns = ['a_neg', 'a_zero', 'a_pos']
by_b = df.groupby(['id', 'b']).count().reset_index().pivot('id', 'b', 'a').fillna(0).astype(int)
by_b.columns = ['b_neg', 'b_zero', 'b_pos']
df_result = by_a.join(by_b).reset_index()
However, I believe that that method is not optimal especially if I have a lot of original columns aside from a and b. Is there a shorter and/or more efficient solution for getting what I want to achieve here? Thanks.

A shorter solution, though still quite in-efficient:
In [11]: df1 = df.set_index("id")
In [12]: g = df1.groupby(level=0)
In [13]: g.apply(lambda x: x.apply(lambda x: x.value_counts())).fillna(0).astype(int).unstack(1)
Out[13]:
a b
-1 0 1 -1 0 1
id
1 1 1 2 1 2 1
2 1 2 1 1 1 2
3 1 2 0 1 1 1
Note: I think you should be aiming for the multi-index columns.
I'm reasonably sure I've seen a trick to remove the apply/value_count/fillna with something cleaner and more efficient, but at the moment it eludes me...

Make Tuples from Specific Pandas Columns

I have a pandas dataframe, e.g.
one two three four five
0 1 2 3 4 5
1 1 1 1 1 1
What I would like is to be able to convert only a select number of columns to a list, such that we obtain:
[[1,2],[1,1]]
This is the rows 0,1, where we are selecting columns one and two.
Similarly if we selected columns one, two, four:
[[1,2,4],[1,1,1]]
Ideally I would like to avoid iteration of rows as it is slow!

You can select just those columns with:
In [11]: df[['one', 'two']]
Out[11]:
one two
0 1 2
1 1 1
and get the list of lists from the underlying numpy array using tolist:
In [12]: df[['one', 'two']].values.tolist()
Out[12]: [[1, 2], [1, 1]]
In [13]: df[['one', 'two', 'four']].values.tolist()
Out[13]: [[1, 2, 4], [1, 1, 1]]
Note: this should never really be necessary unless this is your end game... it's going to be much more efficient to do the work inside pandas or numpy.

So I worked out how to do it.
Firstly we select the columns we would like the values from:
y = x[['one','two']]
This gives us a subset df.
Now we can choose the values:
> y.values
array([[1, 2],
[1, 1]])