I have a dataframe with several numeric columns and their range goes either from 1 to 5 or 1 to 10
I want to create two lists of these columns names this way:
names_1to5 = list of all columns in df with numbers ranging from 1 to 5
names_1to10 = list of all columns in df with numbers from 1 to 10
Example:
IP track batch size type
1 2 3 5 A
9 1 2 8 B
10 5 5 10 C
from the dataframe above:
names_1to5 = ['track', 'batch']
names_1to10 = ['ip', 'size']
I want to use a function that gets a dataframe and perform the above transformation only on columns with numbers within those ranges.
I know that if the column 'max()' is 5 than it's 1to5 same idea when max() is 10
What I already did:
def test(df):
list_1to5 = []
list_1to10 = []
for col in df:
if df[col].max() == 5:
list_1to5.append(col)
else:
list_1to10.append(col)
return list_1to5, list_1to10
I tried the above but it's returning the following error msg:
'>=' not supported between instances of 'float' and 'str'
The type of the columns is 'object' maybe this is the reason. If this is the reason, how can I fix the function without the need to cast these columns to float as there are several, sometimes hundreds of these columns and if I run:
df['column'].max() I get 10 or 5
What's the best way to create this this function?
Use:
string = """alpha IP track batch size
A 1 2 3 5
B 9 1 2 8
C 10 5 5 10"""
temp = [x.split() for x in string.split('\n')]
cols = temp[0]
data = temp[1:]
def test(df):
list_1to5 = []
list_1to10 = []
for col in df.columns:
if df[col].dtype!='O':
if df[col].max() == 5:
list_1to5.append(col)
else:
list_1to10.append(col)
return list_1to5, list_1to10
df = pd.DataFrame(data, columns = cols, dtype=float)
Output:
(['track', 'batch'], ['IP', 'size'])
Related
I have two data frames as shown below df1 and df2. I want to create a third dataframe i.e. df as shown below. What would be the appropriate way?
df1={'id':['a','b','c'],
'val':[1,2,3]}
df1=pd.DataFrame(df)
df1
id val
0 a 1
1 b 2
2 c 3
df2={'yr':['2010','2011','2012'],
'val':[4,5,6]}
df2=pd.DataFrame(df2)
df2
yr val
0 2010 4
1 2011 5
2 2012 6
df={'id':['a','b','c'],
'val':[1,2,3],
'2010':[4,8,12],
'2011':[5,10,15],
'2012':[6,12,18]}
df=pd.DataFrame(df)
df
id val 2010 2011 2012
0 a 1 4 5 6
1 b 2 8 10 12
2 c 3 12 15 18
I can basically convert df1 and df2 as 1 by n matrices and get n by n result and assign it back to the df1. But is there any easy pandas way?
TL;DR
We can do it in one line like this:
df1.join(df1.val.apply(lambda x: x * df2.set_index('yr').val))
or like this:
df1.join(df1.set_index('id') # df2.set_index('yr').T, on='id')
Done.
The long story
Let's see what's going on here.
To find the output of multiplication of each df1.val by values in df2.val we use apply:
df1['val'].apply(lambda x: x * df2.val)
The function inside will obtain df1.vals one by one and multiply each by df2.val element-wise (see broadcasting for details if needed). As far as df2.val is a pandas sequence, the output is a data frame with indexes df1.val.index and columns df2.val.index. By df2.set_index('yr') we force years to be indexes before multiplication so they will become column names in the output.
DataFrame.join is joining frames index-on-index by default. So due to identical indexes of df1 and the multiplication output, we can apply df1.join( <the output of multiplication> ) as is.
At the end we get the desired matrix with indexes df1.index and columns id, val, *df2['yr'].
The second variant with # operator is actually the same. The main difference is that we multiply 2-dimentional frames instead of series. These are the vertical and horizontal vectors, respectively. So the matrix multiplication will produce a frame with indexes df1.id and columns df2.yr and element-wise multiplication as values. At the end we connect df1 with the output on identical id column and index respectively.
This works for me:
df2 = df2.T
new_df = pd.DataFrame(np.outer(df1['val'],df2.iloc[1:]))
df = pd.concat([df1, new_df], axis=1)
df.columns = ['id', 'val', '2010', '2011', '2012']
df
The output I get:
id val 2010 2011 2012
0 a 1 4 5 6
1 b 2 8 10 12
2 c 3 12 15 18
Your question is a bit vague. But I suppose you want to do something like that:
df = pd.concat([df1, df2], axis=1)
I usually use "${:,.2f}". format(prices) to round numbers before commas, but what I'm looking for is different, I want to change values numbers to group them and reference them by mode:
Let say I have this list:
0 34,123.45
1 34,456.78
2 34,567.89
3 33,222.22
4 30,123.45
And the replace function will turn the list to:
0 34,500.00
1 34,500.00
2 34,500.00
3 33,200.00
4 30,100.00
Like this when I use stats.mode(prices_rounded) it will show as a result:
Mode Value = 34500.00
Mode Count = 3
Is there a conversion function already available that does the job? I did search for days without luck...
EDIT - WORKING CODE:
#create list
df3 = df_array
print('########## df3: ',df3)
#convert to float
df4 = df3.astype(float)
print('########## df4: ',df4)
#convert list to string
#df5 = ''.join(map(str, df4))
#print('########## df5: ',df5)
#round values
df6 = np.round(df4 /100) * 100
print('######df6',df6)
#get mode stats
df7 = stats.mode(df6)
print('######df7',df7)
#get mode value
df8 = df7[0][0]
print('######df8',df8)
#convert to integer
df9 = int(df8)
print('######df9',df9)
This is exactly what I wanted, thanks!
You can use:
>>> sr
0 34123.45 # <- why 34500.00?
1 34456.78
2 34567.89 # <- why 34500.00?
3 33222.22
4 30123.45
dtype: float64
>>> np.round(sr / 100) * 100
0 34100.0
1 34500.0
2 34600.0
3 33200.0
4 30100.0
dtype: float64
How do I compare the value of the first row in col b and the last row in col b from grouping by col a, without using the groupby function? Because groupby function is very slow for a large dataset.
a = [1,1,1,1,1,1,1,2,2,2,2,2,2,2,3,3,3,3,3,3,3]
b = [1,0,0,0,0,0,7,8,0,0,0,0,0,4,1,0,0,0,0,0,1]
Return two lists: one has the group names from col a where the last value is larger than the first value, etc.
larger_or_equal = [1,3]
smaller = [2]
All numpy
a = np.array([1,1,1,1,1,1,1,2,2,2,2,2,2,2,3,3,3,3,3,3,3])
b = np.array([1,0,0,0,0,0,7,8,0,0,0,0,0,4,1,0,0,0,0,0,1])
w = np.where(a[1:] != a[:-1])[0] # find the edges
e = np.append(w, len(a) - 1) # define the end pos
s = np.append(0, w + 1) # define start pos
# slice end pos with boolean array. then slice groups with end postions.
# I could also have used start positions.
a[e[b[e] >= b[s]]]
a[e[b[e] < b[s]]]
[1 3]
[2]
Here is a solution without groupby. The idea is to shift column a to detect group changes:
df[df['a'].shift() != df['a']]
a b
0 1 1
7 2 8
14 3 1
df[df['a'].shift(-1) != df['a']]
a b
6 1 7
13 2 4
20 3 1
We will compare the column b in those two dataframes. We simply need to reset the index for the pandas comparison to work:
first = df[df['a'].shift() != df['a']].reset_index(drop=True)
last = df[df['a'].shift(-1) != df['a']].reset_index(drop=True)
first.loc[last['b'] >= first['b'], 'a'].values
array([1, 3])
Then do the same with < to get the other groups. Or do a set difference.
As I wrote in the comments, groupby(sort=False) might well be faster depending on your dataset.
I am trying to calculate a percent change between 2 numbers in one column when a signal from another column is triggered.
The trigger can be found with np.where() but what I am having trouble with is the percent change. .pct_change does not work because if you .pct_change(-5) you get 16.03/20.35 and I want the number the opposite way 20.35/16.03. See table below. I have tried returning the array from the index in the np.where and adding it to an .iloc from the 'Close' column but it says I can't use that array to get an .iloc position. Can anyone help me solve this problem. Thank you.
IdxNum | Close | Signal (1s)
==============================
0 21.45 0
1 21.41 0
2 21.52 0
3 21.71 0
4 20.8 0
5 20.35 0
6 20.44 0
7 16.99 0
8 17.02 0
9 16.69 0
10 16.03 1<< 26.9% <<< 20.35/16.03-1 (df.Close[5]/df.Close[10]-1)
11 15.67 0
12 15.6 0
You can try this code block:
#Create DataFrame
df = pd.DataFrame({'IdxNum':range(13),
'Close':[21.45,21.41,21.52,21.71,20.8,20.35,20.44,16.99,17.02,16.69,16.03,15.67,15.6],
'Signal':[0] * 13})
df.ix[10,'Signal']=1
#Create a function that calculates the reqd diff
def cal_diff(row):
if(row['Signal']==1):
signal_index = int(row['IdxNum'])
row['diff'] = df.Close[signal_index-5]/df.Close[signal_index]-1
return row
#Create a column and apply that difference
df['diff'] = 0
df = df.apply(lambda x:cal_diff(x),axis=1)
In case you don't have IdxNum column, you can use the index to calculate difference
#Create DataFrame
df = pd.DataFrame({
'Close':[21.45,21.41,21.52,21.71,20.8,20.35,20.44,16.99,17.02,16.69,16.03,15.67,15.6],
'Signal':[0] * 13})
df.ix[10,'Signal']=1
#Calculate the reqd difference
df['diff'] = 0
signal_index = df[df['Signal']==1].index[0]
df.ix[signal_index,'diff'] = df.Close[signal_index-5]/df.Close[signal_index]-1
Suppose I have a set of measurements that were obtained by varying two parameters, knob_b and knob_2 (in practice there are a lot more):
data = np.empty((6,3), dtype=np.float)
data[:,0] = [3,4,5,3,4,5]
data[:,1] = [1,1,1,2,2,2]
data[:,2] = np.random.random(6)
df = pd.DataFrame(data, columns=['knob_1', 'knob_2', 'signal'])
i.e., df is
knob_1 knob_2 signal
0 3 1 0.076571
1 4 1 0.488965
2 5 1 0.506059
3 3 2 0.415414
4 4 2 0.771212
5 5 2 0.502188
Now, considering each parameter on its own, I want to find the minimum value that was measured for each setting of this parameter (ignoring the settings of all other parameters). The pedestrian way of doing this is:
new_index = []
new_data = []
for param in df.columns:
if param == 'signal':
continue
group = df.groupby(param)['signal'].min()
for (k,v) in group.items():
new_index.append((param, k))
new_data.append(v)
new_index = pd.MultiIndex.from_tuples(new_index,
names=('parameter', 'value'))
df2 = pd.Series(index=new_index, data=new_data)
resulting df2 being:
parameter value
knob_1 3 0.495674
4 0.277030
5 0.398806
knob_2 1 0.485933
2 0.277030
dtype: float64
Is there a better way to do this, in particular to get rid of the inner loop?
It seems to me that the result of the df.groupby operation already has everything I need - if only there was a way to somehow create a MultiIndex from it without going through the list of tuples.
Use the keys argument of pd.concat():
pd.concat([df.groupby('knob_1')['signal'].min(),
df.groupby('knob_2')['signal'].min()],
keys=['knob_1', 'knob_2'],
names=['parameter', 'value'])