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Compute difference between values in dataframe column

i have this dataframe:
a b c d
4 7 5 12
3 8 2 8
1 9 3 5
9 2 6 4
i want the column 'd' to become the difference between n-value of column a and n+1 value of column 'a'.
I tried this but it doesn't run:
for i in data.index-1:
data.iloc[i]['d']=data.iloc[i]['a']-data.iloc[i+1]['a']
can anyone help me?

Basically what you want is diff.
df = pd.DataFrame.from_dict({"a":[4,3,1,9]})
df["d"] = df["a"].diff(periods=-1)
print(df)
Output
a d
0 4 1.0
1 3 2.0
2 1 -8.0
3 9 NaN

lets try simple way:
df=pd.DataFrame.from_dict({'a':[2,4,8,15]})
diff=[]
for i in range(len(df)-1):
diff.append(df['a'][i+1]-df['a'][i])
diff.append(np.nan)
df['d']=diff
print(df)
a d
0 2 2.0
1 4 4.0
2 8 7.0
3 15 NaN

Finding mean of specific column and keep all rows that have specific mean values

I have this dataframe.
from pandas import DataFrame
import pandas as pd
df = pd.DataFrame({'name': ['A','D','M','T','B','C','D','E','A','L'],
'id': [1,1,1,2,2,3,3,3,3,5],
'rate': [3.5,4.5,2.0,5.0,4.0,1.5,2.0,2.0,1.0,5.0]})
>> df
name id rate
0 A 1 3.5
1 D 1 4.5
2 M 1 2.0
3 T 2 5.0
4 B 2 4.0
5 C 3 1.5
6 D 3 2.0
7 E 3 2.0
8 A 3 1.0
9 L 5 5.0
df = df.groupby('id')['rate'].mean()
what i want is this:
1) find mean of every 'id'.
2) give the number of ids (length) which has mean >= 3.
3) give back all rows of dataframe (where mean of any id >= 3.
Expected output:
Number of ids (length) where mean >= 3: 3
>> dataframe where (mean(id) >=3)
>>df
name id rate
0 A 1 3.0
1 D 1 4.0
2 M 1 2.0
3 T 2 5.0
4 B 2 4.0
5 L 5 5.0

Use GroupBy.transform for means by all groups with same size like original DataFrame, so possible filter by boolean indexing:
df = df[df.groupby('id')['rate'].transform('mean') >=3]
print (df)
name id rate
0 A 1 3.5
1 D 1 4.5
2 M 1 2.0
3 T 2 5.0
4 B 2 4.0
9 L 5 5.0
Detail:
print (df.groupby('id')['rate'].transform('mean'))
0 3.333333
1 3.333333
2 3.333333
3 4.500000
4 4.500000
5 1.625000
6 1.625000
7 1.625000
8 1.625000
9 5.000000
Name: rate, dtype: float64
Alternative solution with DataFrameGroupBy.filter:
df = df.groupby('id').filter(lambda x: x['rate'].mean() >=3)

Backfilling columns by groups in Pandas

I have a csv like
A,B,C,D
1,2,,
1,2,30,100
1,2,40,100
4,5,,
4,5,60,200
4,5,70,200
8,9,,
In row 1 and row 4 C value is missing (NaN). I want to take their value from row 2 and 5 respectively. (First occurrence of same A,B value).
If no matching row is found, just put 0 (like in last line)
Expected op:
A,B,C,D
1,2,30,
1,2,30,100
1,2,40,100
4,5,60,
4,5,60,200
4,5,70,200
8,9,0,
using fillna I found bfill: use NEXT valid observation to fill gap but the NEXT observation has to be taken logically (looking at col A,B values) and not just the upcoming C column value

You'll have to call df.groupby on A and B first and then apply the bfill function:
In [501]: df.C = df.groupby(['A', 'B']).apply(lambda x: x.C.bfill()).reset_index(drop=True)
In [502]: df
Out[502]:
A B C D
0 1 2 30 NaN
1 1 2 30 100.0
2 1 2 40 100.0
3 4 5 60 NaN
4 4 5 60 200.0
5 4 5 70 200.0
6 8 9 0 NaN
You can also group and then call dfGroupBy.bfill directly (I think this would be faster):
In [508]: df.C = df.groupby(['A', 'B']).C.bfill().fillna(0).astype(int); df
Out[508]:
A B C D
0 1 2 30 NaN
1 1 2 30 100.0
2 1 2 40 100.0
3 4 5 60 NaN
4 4 5 60 200.0
5 4 5 70 200.0
6 8 9 0 NaN
If you wish to get rid of NaNs in D, you could do:
df.D.fillna('', inplace=True)

Find observations in which both columns are NaN and replace them with 0 in pandas DataFrame

Here is a dataframe
a b c d
nan nan 3 5
nan 1 2 3
1 nan 4 5
2 3 7 9
nan nan 2 3
I want to replace the observations in both columns 'a' and 'b' where both of them are NaNs with 0s. Rows 2 and 5 in columns 'a' and 'b' have both both NaN, so I want to replace only those rows with 0's in those matching NaN columns.
so my output must be
a b c d
0 0 3 5
nan 1 2 3
1 nan 4 5
2 3 7 9
0 0 2 3

There might be a easier builtin function in Pandas, but this one should work.
df[['a', 'b']] = df.ix[ (np.isnan(df.a)) & (np.isnan(df.b)), ['a', 'b'] ].fillna(0)
Actually the solution from #Psidom much easier to read.

You can create a boolean series based on the conditions on columns a/b, and then use loc to modify corresponding columns and rows:
df.loc[df[['a','b']].isnull().all(1), ['a','b']] = 0
df
# a b c d
#0 0.0 0.0 3 5
#1 NaN 1.0 2 3
#2 1.0 NaN 4 5
#3 2.0 3.0 7 9
#4 0.0 0.0 2 3
Or:
df.loc[df.a.isnull() & df.b.isnull(), ['a','b']] = 0

Why Python Pandas append to DataFrame like this?

I want to add l in column 'A' but it creates a new column and adds l to the last one. Why is it happening? And how can I make what I want?
import pandas as pd
l=[1,2,3]
df = pd.DataFrame(columns =['A'])
df = df.append(l, ignore_index=True)
df = df.append(l, ignore_index=True)
print(df)
A 0
0 NaN 1.0
1 NaN 2.0
2 NaN 3.0
3 NaN 1.0
4 NaN 2.0
5 NaN 3.0

Edited
Is this what you want to do:
In[6]:df=df.A.append(pd.Series(l)).reset_index().drop('index',1).rename(columns={0:'A'})
In[7]:df
Out[7]:
A
0 1
1 2
2 3
Then you can add any list of different length.
Suppose:
a=[9,8,7,6,5]
In[11]:df=df.A.append(pd.Series(a)).reset_index().drop('index',1).rename(columns={0:'A'})
In[12]:df
Out[12]:
A
0 1
1 2
2 3
3 9
4 8
5 7
6 6
7 5
Previously
are you looking for this :
df=pd.DataFrame(l,columns=['A'])
df
Out[5]:
A
0 1
1 2
2 3

You can just pass a dictionary in the dataframe constructor, that if I understand your question correctly.
l = [1,2,3]
df = pd.DataFrame({'A': l})
df
A
0 1
1 2
2 3