Python FPDF fill color of a cell based on an IF statement

Team,

How do I change the cell background color of a table. For example, I want to change the cell background of column B items if it is "No" using FPDF and IF statement or other methods. That is if a No is in the cell, the "No" cell should be colored.

Column A	Column B
Yes	No
Yes	No

    if df.Column B.any() == "No":
        pdf.set_fill_color(193, 229, 252)
        pdf.cell(30, 10, 'No', 1, 1, 'C', True)

But it did not work.

Solution

Besides the syntax error in df.Column B, df["Column B"].any() will always returns True.

You can try using iterrows from pandas :

from fpdf import FPDF

pdf = FPDF()
pdf.add_page()

pdf.set_fill_color(255, 255, 255) #make a white-bg
pdf.set_font("Arial", "B", 12) # <-- ajudst here if needed
for col in df.columns:
    pdf.cell(30, 10, col, 1, 0, "C", 1)
pdf.ln(10) #a newline

for _, row in df.iterrows():
    pdf.set_font("Arial", "", 10)  # <-- ajudst here if needed
    pdf.cell(30, 10, row["Column A"], 1, 0, "C")

    if row["Column B"] == "No":
        pdf.set_fill_color(193, 229, 252) #make a blue-bg
        pdf.cell(30, 10, row["Column B"], 1, 0, "C", 1)
    else:
        pdf.cell(30, 10, row["Column B"], 1, 0, "C")

    pdf.ln(10) #a newline

pdf.output("output.pdf", "F")

Output (.pdf) :

Equivalent with Styler (in Jupyter) :

(
    df.style
     .applymap(lambda x: "background-color: #C1E5FC"
               if x == "No" else "", subset=["Column B"])
)

UPDATE :

Actually I have four columns. There are 2 columns before column A. How do I re-write for pdf.cell(30, 10, row["Column A"], 1) for each column preceeding column B ?

for _, row in df.iterrows():
    pdf.set_font("Arial", "", 10)
    pdf.cell(30, 10, row["Column 1"], 1, 0, "C") # <-- add this line
    pdf.cell(30, 10, row["Column 2"], 1, 0, "C") # <-- add this line
    pdf.cell(30, 10, row["Column A"], 1, 0, "C")

    if row["Column B"] == "No":
        pdf.set_fill_color(193, 229, 252) #make a blue-bg
        pdf.cell(30, 10, row["Column B"], 1, 0, "C", 1)
    else:
        pdf.cell(30, 10, row["Column B"], 1, 0, "C")

    pdf.ln(10) #a newline

pdf.output("output.pdf", "F")

Output :

Input used :

  Column 1 Column 2 Column A Column B
0        A        X      foo      Yes
1        B        Y      bar       No
2        C        Z      baz      Yes