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How to write excel-like formulas in Python

I developed a model in excel and now I am trying to write this model totally in python. But for calculation, I cannot think of another way to do the calculations for all the cells than with a loop, which makes the processing time too long. I want to know if there is another way to write the code and therefore reduce the time to process all the data.
peak_start = pd.to_timedelta('08:00:00')
peak_end = pd.to_timedelta('20:00:00')
for i in df_open_position.columns[3:10]:
df_open_position[i] = df_open_position['Index'][:8760].apply(lambda x: pf_targets.loc[(pf_targets['Target Startdatum'] <= df_open_position['Liefertag'][x]) & (pf_targets['Target Enddatum'] >= df_open_position['Liefertag'][x]) & (pf_targets['Produkt'].str.contains('Base')), 'Menge'].sum() \
+ pf_targets.loc[(df_open_position['Liefertag'][x].weekday() >= 5) & (peak_start <= df_open_position['Zeit'][x] <= peak_end) & (pf_targets['Produkt'].str.contains('Peak')), 'Menge'].sum() \
+ pf_swing_targets['Menge [MW]'][x] \
- pf_deals.loc[(pf_deals['Lieferbeginn'] <= df_open_position['Liefertag'][x]) & (pf_deals['Lieferende'] >= df_open_position['Liefertag'][x]) & (pf_deals[' Kaufrichtung'] == "Kauf") & (pf_deals[' Datum'] <= df_open_position['Liefertag'][x]) & (pf_deals[' Produkt'].str.contains('Base')), ' Menge' ].sum() \
- pf_deals.loc[(df_open_position['Liefertag'][x].weekday() >= 5) & (peak_start <= df_open_position['Zeit'][x] <= peak_end) & (pf_deals[' Produkt'].str.contains('Peak')) & (pf_deals['Lieferbeginn'] <= df_open_position['Liefertag'][x]) & (pf_deals['Lieferende'] >= df_open_position['Liefertag'][x]) & (pf_deals[' Kaufrichtung'] == "Kauf") & (pf_deals[' Datum'] <= df_open_position['Liefertag'][x]), ' Menge'].sum() \
+ pf_deals.loc[(pf_deals['Lieferbeginn'] <= df_open_position['Liefertag'][x]) & (pf_deals['Lieferende'] >= df_open_position['Liefertag'][x]) & (pf_deals[' Kaufrichtung'] == "Verkauf") & (pf_deals[' Datum'] <= df_open_position['Liefertag'][x]) & (pf_deals[' Produkt'].str.contains('Base')), ' Menge' ].sum() \
+ pf_deals.loc[(df_open_position['Liefertag'][x].weekday() >= 5) & (peak_start <= df_open_position['Zeit'][x] <= peak_end) & (pf_deals[' Produkt'].str.contains('Peak')) & (pf_deals['Lieferbeginn'] <= df_open_position['Liefertag'][x]) & (pf_deals['Lieferende'] >= df_open_position['Liefertag'][x]) & (pf_deals[' Kaufrichtung'] == "Verkauf") & (pf_deals[' Datum'] <= df_open_position['Liefertag'][x]), ' Menge'].sum())
Here is the code that I'm using in my excel file:
=IF(E$1="";"";
SUMIFS(Targets!$G:$G;Targets!$I:$I;"<="&$A3;Targets!$J:$J;">="&$A3;Targets!$F:$F;"*"&"Base")
+SUMIFS(Targets!$G:$G;Targets!$I:$I;"<="&$A3;Targets!$J:$J;">="&$A3;Targets!$F:$F;"TS")
+IF(AND(WORKDAY($A3-1;1)=$A3;$D3="peak");SUMIFS(Targets!$G:$G;Targets!$I:$I;"<="&$A3;Targets!$J:$J;">="&$A3;Targets!$F:$F;"*"&"Peak");)
-SUMIFS(Deals!$K:$K;Deals!$R:$R;"<="&$A3;Deals!$S:$S;">="&$A3;Deals!$J:$J;"Kauf";Deals!$I:$I;"<="&'PF 22'!E$1;Deals!$G:$G;"*"&"Base")
-IF(AND(WORKDAY($A3-1;1)=$A3;$D3="peak");SUMIFS(Deals!$K:$K;Deals!$R:$R;"<="&$A3;Deals!$S:$S;">="&$A3;Deals!$J:$J;"Kauf";Deals!$I:$I;"<="&'PF 22'!E$1;Deals!$G:$G;"*"&"Peak");)
+ SUMIFS(Deals!$K:$K;Deals!$R:$R;"<="&$A3;Deals!$S:$S;">="&$A3;Deals!$J:$J;"Verkauf";Deals!$I:$I;"<="&'PF 22'!E$1;Deals!$G:$G;"*"&"Base")
+IF(AND(WORKDAY($A3-1;1)=$A3;$D3="peak");SUMIFS(Deals!$K:$K;Deals!$R:$R;"<="&$A3;Deals!$S:$S;">="&$A3;Deals!$J:$J;"Verkauf";Deals!$I:$I;"<="&'PF 22'!E$1;Deals!$G:$G;"*"&"Peak");)
+$Z3)
I hope with all the information provided someone can help me.

Python: Remove substrings from string starting and ending with certain words

I have a string of this kind of type:
import re
s = 'T [90] Call: proof(([temp(p∨q∨r), p, ¬q], [])⊢q, _41226, _41228, proof{ \
-5: ["assumptions([p,¬q,p∨q∨r])","premisses_origin([q,¬q])","premisses_no_origin([])","premisses_exc_origin([])","conclusion([q∧ ¬q])","rule([∧I])","d0(([temp(p∨q∨r),p,¬q],[])⊢(q∧ ¬q))","d1(([temp(p∨q∨r),p,¬q],[])⊢q∧([temp(p∨q∨r),p,¬q],[q])⊢(¬q))","step(9)"] \
\
-4: ["assumptions([p,¬q,p∨q∨r])","premisses_origin([¬q,q∧ ¬q])","premisses_no_origin([])","premisses_exc_origin([¬q])","conclusion([q])","rule([¬E])","d0(([temp(p∨q∨r),p],[])⊢q)","d1(([temp(p∨q∨r),p,¬q],[])⊢(q∧ ¬q))","step(8)"]\
\
-3: ["assumptions([p,p∨q∨r])","premisses_origin([q])","premisses_no_origin([])","premisses_exc_origin([])","conclusion([q∨p∨r])","rule([∨I])","d0(([temp(p∨q∨r),p],[])⊢(q∨p∨r))","d1(([temp(p∨q∨r),p],[])⊢q)","step(6)"]\
\
-2: ["assumptions([p,p∨q∨r])","premisses_origin([q∨p∨r])","premisses_no_origin([])","premisses_exc_origin([p])","conclusion([p→q∨p∨r])","rule([→I])","d0(([temp(p∨q∨r)],[])⊢(p→q∨p∨r))","d1(([temp(p∨q∨r),p],[])⊢(q∨p∨r))","step(5)"]\
\
-1: ["assumptions([p∨q∨r])","premisses_origin([p→q∨p∨r,q∨r→q∨p∨r])","premisses_no_origin([])","premisses_exc_origin([])","conclusion([(p→q∨p∨r)∧(q∨r→q∨p∨r)])","rule([∧I])","d0(([temp(p∨q∨r)],[])⊢((p→q∨p∨r)∧(q∨r→q∨p∨r)))","d1(([temp(p∨q∨r)],[])⊢(p→q∨p∨r)∧([temp(p∨q∨r)],[p→q∨p∨r])⊢(q∨r→q∨p∨r))","step(3)"]\
\
0: ["assumptions([p∨q∨r])","premisses_origin([p∨q∨r,(p→q∨p∨r)∧(q∨r→q∨p∨r)])","premisses_no_origin([])","premisses_exc_origin([])","conclusion([q∨p∨r])","rule([∨E])","d0(([p∨q∨r],[])⊢(q∨p∨r))","d1(([temp(p∨q∨r)],[])⊢((p→q∨p∨r)∧(q∨r→q∨p∨r)))","step(2)"]\
\
1: ["p∨q∨r","step(1)"]\
\
2: ["p","step(4)"]\
\
3: ["¬q","step(7)"]\
\
}, _41204, _41234)\
T [90] Fail: proof(([temp(p∨q∨r), p, ¬q], [])⊢q, _41226, _41228, proof{\
-5: ["assumptions([p,¬q,p∨q∨r])","premisses_origin([q,¬q])","premisses_no_origin([])","premisses_exc_origin([])","conclusion([q∧ ¬q])","rule([∧I])","d0(([temp(p∨q∨r),p,¬q],[])⊢(q∧ ¬q))","d1(([temp(p∨q∨r),p,¬q],[])⊢q∧([temp(p∨q∨r),p,¬q],[q])⊢(¬q))","step(9)"]\
\
-4: ["assumptions([p,¬q,p∨q∨r])","premisses_origin([¬q,q∧ ¬q])","premisses_no_origin([])","premisses_exc_origin([¬q])","conclusion([q])","rule([¬E])","d0(([temp(p∨q∨r),p],[])⊢q)","d1(([temp(p∨q∨r),p,¬q],[])⊢(q∧ ¬q))","step(8)"]\
\
-3: ["assumptions([p,p∨q∨r])","premisses_origin([q])","premisses_no_origin([])","premisses_exc_origin([])","conclusion([q∨p∨r])","rule([∨I])","d0(([temp(p∨q∨r),p],[])⊢(q∨p∨r))","d1(([temp(p∨q∨r),p],[])⊢q)","step(6)"]\
\
-2: ["assumptions([p,p∨q∨r])","premisses_origin([q∨p∨r])","premisses_no_origin([])","premisses_exc_origin([p])","conclusion([p→q∨p∨r])","rule([→I])","d0(([temp(p∨q∨r)],[])⊢(p→q∨p∨r))","d1(([temp(p∨q∨r),p],[])⊢(q∨p∨r))","step(5)"]\
\
-1: ["assumptions([p∨q∨r])","premisses_origin([p→q∨p∨r,q∨r→q∨p∨r])","premisses_no_origin([])","premisses_exc_origin([])","conclusion([(p→q∨p∨r)∧(q∨r→q∨p∨r)])","rule([∧I])","d0(([temp(p∨q∨r)],[])⊢((p→q∨p∨r)∧(q∨r→q∨p∨r)))","d1(([temp(p∨q∨r)],[])⊢(p→q∨p∨r)∧([temp(p∨q∨r)],[p→q∨p∨r])⊢(q∨r→q∨p∨r))","step(3)"]\
\
0: ["assumptions([p∨q∨r])","premisses_origin([p∨q∨r,(p→q∨p∨r)∧(q∨r→q∨p∨r)])","premisses_no_origin([])","premisses_exc_origin([])","conclusion([q∨p∨r])","rule([∨E])","d0(([p∨q∨r],[])⊢(q∨p∨r))","d1(([temp(p∨q∨r)],[])⊢((p→q∨p∨r)∧(q∨r→q∨p∨r)))","step(2)"]\
\
1: ["p∨q∨r","step(1)"]\
\
2: ["p","step(4)"]\
\
3: ["¬q","step(7)"]\
\
}, _41204, _41234)\
T [81] Fail: proof(([temp(p∨q∨r), p, ¬q], [])⊢q∧([temp(p∨q∨r), p, ¬q], [q])⊢(¬q), _38484, _38486, proof{\
-5: ["assumptions([p,¬q,p∨q∨r])","premisses_origin([q,¬q])","premisses_no_origin([])","premisses_exc_origin([])","conclusion([q∧ ¬q])","rule([∧I])","d0(([temp(p∨q∨r),p,¬q],[])⊢(q∧ ¬q))","d1(([temp(p∨q∨r),p,¬q],[])⊢q∧([temp(p∨q∨r),p,¬q],[q])⊢(¬q))","step(9)"]\
\
-4: ["assumptions([p,¬q,p∨q∨r])","premisses_origin([¬q,q∧ ¬q])","premisses_no_origin([])","premisses_exc_origin([¬q])","conclusion([q])","rule([¬E])","d0(([temp(p∨q∨r),p],[])⊢q)","d1(([temp(p∨q∨r),p,¬q],[])⊢(q∧ ¬q))","step(8)"]\
\
-3: ["assumptions([p,p∨q∨r])","premisses_origin([q])","premisses_no_origin([])","premisses_exc_origin([])","conclusion([q∨p∨r])","rule([∨I])","d0(([temp(p∨q∨r),p],[])⊢(q∨p∨r))","d1(([temp(p∨q∨r),p],[])⊢q)","step(6)"]\
\
-2: ["assumptions([p,p∨q∨r])","premisses_origin([q∨p∨r])","premisses_no_origin([])","premisses_exc_origin([p])","conclusion([p→q∨p∨r])","rule([→I])","d0(([temp(p∨q∨r)],[])⊢(p→q∨p∨r))","d1(([temp(p∨q∨r),p],[])⊢(q∨p∨r))","step(5)"]\
\
-1: ["assumptions([p∨q∨r])","premisses_origin([p→q∨p∨r,q∨r→q∨p∨r])","premisses_no_origin([])","premisses_exc_origin([])","conclusion([(p→q∨p∨r)∧(q∨r→q∨p∨r)])","rule([∧I])","d0(([temp(p∨q∨r)],[])⊢((p→q∨p∨r)∧(q∨r→q∨p∨r)))","d1(([temp(p∨q∨r)],[])⊢(p→q∨p∨r)∧([temp(p∨q∨r)],[p→q∨p∨r])⊢(q∨r→q∨p∨r))","step(3)"]\
\
0: ["assumptions([p∨q∨r])","premisses_origin([p∨q∨r,(p→q∨p∨r)∧(q∨r→q∨p∨r)])","premisses_no_origin([])","premisses_exc_origin([])","conclusion([q∨p∨r])","rule([∨E])","d0(([p∨q∨r],[])⊢(q∨p∨r))","d1(([temp(p∨q∨r)],[])⊢((p→q∨p∨r)∧(q∨r→q∨p∨r)))","step(2)"]\
\
1: ["p∨q∨r","step(1)"]\
\
2: ["p","step(4)"]\
\
3: ["¬q","step(7)"]\
\
}, _31664, 0)'
my aim now is to remove all substrings starting with proof{ and ending with } such that the aim output is something like that:
result = 'T [90] Call: proof(([temp(p∨q∨r), p, ¬q], [])⊢q, _41226, _41228, , _41204, _41234)\
T [90] Fail: proof(([temp(p∨q∨r), p, ¬q], [])⊢q, _41226, _41228, , _41204, _41234)\
T [81] Fail: proof(([temp(p∨q∨r), p, ¬q], [])⊢q∧([temp(p∨q∨r), p, ¬q], [q])⊢(¬q), _38484, _38486, , _31664, 0)'
Based on a similar question I tried something like that:
start = re.escape("proof{")
end = re.escape("}")
result = re.search('%s(.*)%s' % (start, end), s).group(1)
but it doesn't do what I want.

I think I found it:
re.sub("proof{.*?}",'', s)

what is the correct method to print this image?

I'm trying to print the following image via python
print("""
____
.\ /
|\\ //\
/ \\// \
/ / \ \
/ / \ \
/ / \ \
/ /______^ \ \
/ ________\ \ \
/ / \ \
/\\ / \ //\
/__\\_\ /_//__\
""")
input()
output
____
.\ /
|\ // / \// / / \ / / \ / / \ \
/ /______^ \ / ________\ \ / / \ /\ / \ ///__\_\ /_//__
hope someone can help me solve this problem

Backslashes escape the newlines, change it to a raw string with r"...":
print(r"""
____
.\ /
|\\ //\
/ \\// \
/ / \ \
/ / \ \
/ / \ \
/ /______^ \ \
/ ________\ \ \
/ / \ \
/\\ / \ //\
/__\\_\ /_//__\
""")
input()

What's wrong on my conditions ? Using the np.where statement to flag my pandas dataframes

The function i am using is keep giving the red filter condition where not applied.
Here the function i am using:
tolerance = 5
def rag(data):
red_filter = ((data.SHIPMENT_MOT_x == 'VESSEL') & \
((data.latedeliverydate + pd.to_timedelta(tolerance,unit='D')) < data.m6p)) | \
((data.SHIPMENT_MOT_x == 'AIR') & (data.latedeliverydate < data.m6p))
green_filter = (data.SHIPMENT_MOT_x == 'VESSEL') & \
(data.M6_proposed == data.m6p) & \
((data.latedeliverydate + pd.to_timedelta(tolerance,unit='D')) >= data.m6p) | \
((data.SHIPMENT_MOT_x == 'AIR') & (data.latedeliverydate >= data.m6p))
amber_filter = (data.SHIPMENT_MOT_x == 'VESSEL') & \
(data.M6_proposed != data.m6p) & \
((data.latedeliverydate + pd.to_timedelta(tolerance,unit='D')) >= data.m6p) | \
((data.SHIPMENT_MOT_x == 'AIR') & (data.latedeliverydate >= data.m6p))
data['RAG'] = np.where(green_filter, 'G', np.where(amber_filter, 'A', np.where(red_filter, 'R', '')))

Here is the solution if you guys are interested.
np.where is useful but would not recommend when there are multiple conditions
def pmm_rag(data):
if ((data.MOT== 'VESSEL') & ((data.m0p + pd.to_timedelta(tolerance,unit='D')) < data.m6p)) | ((data.SHIPMENT_MOT_x == 'AIR') & (data.m0p < data.m6p)):
return 'R'
elif (data.MOT== 'VESSEL') & (data.M6_proposed == data.m6p) & ((data.m0p + pd.to_timedelta(tolerance,unit='D')) >= data.m6p) | ((data.MOT== 'AIR') & (data.m0p >= data.m6p)):
return 'G'
elif (data.MOT== 'VESSEL') & (data.M6_proposed != data.m6p) & ((data.m0p + pd.to_timedelta(tolerance,unit='D')) >= data.m6p) | ((data.MOT== 'AIR') & (data.m0p >= data.m6p)):
return 'A'
else:
return ''

Printing a text tree in python

How would you print a tree in python so it looks like the following:
/\
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
The height should be changeable.

def tree(n):
treeStr = ""
for i in range(0,n):
level = " "*(n-i) + "/" + " "*(2*i) + "\\"
treeStr += level.center(n+1) + "\n"
print(treeStr)
tree(10);