我正在尝试从我创建的字典中制作多列条形图,但不确定如何正确地绘制它
我曾尝试使用熊猫图书馆,但收效甚微
if x==0:
IR10_cnt=IR10_cnt+1
f_cond_r10.append(a)
f_pow_r10.append(b)
f_mom_r10.append(c)
f_ct[x]=[Counter(f_cond_r10), Counter(f_pow_r10), Counter(f_mom_r10)]
f_ct[x]=[dict(Counter(f_cond_r10)),dict(Counter(f_pow_r10)), dict(Counter(f_mom_r10))]
elif x==1:
IR11_cnt=IR11_cnt+1
f_cond_r11.append(a)
f_pow_r11.append(b)
f_mom_r11.append(c)
f_ct[x]=[dict(Counter(f_cond_r11)),dict(Counter(f_pow_r11)), dict(Counter(f_mom_r11))]
elif x==2:
IR12_cnt=IR12_cnt+1
f_cond_r12.append(a)
f_pow_r12.append(b)
f_mom_r12.append(c)
f_ct[x]=[dict(Counter(f_cond_r12)),dict(Counter(f_pow_r12)), dict(Counter(f_mom_r12))]
elif x==3:
IR13_cnt=IR13_cnt+1
f_cond_r13.append(a)
f_pow_r13.append(b)
f_mom_r13.append(c)
f_ct[x]=[dict(Counter(f_cond_r13)),dict(Counter(f_pow_r13)), dict(Counter(f_mom_r13))]
elif x==4:
IR14_cnt=IR14_cnt+1
f_cond_r14.append(a)
f_pow_r14.append(b)
f_mom_r14.append(c)
f_ct[x]=[dict(Counter(f_cond_r14)),dict(Counter(f_pow_r14)), dict(Counter(f_mom_r14))]
elif x==5:
IR15_cnt=IR15_cnt+1
f_cond_r15.append(a)
f_pow_r15.append(b)
f_mom_r15.append(c)
f_ct[x]=[dict(Counter(f_cond_r15)),dict(Counter(f_pow_r15)), dict(Counter(f_mom_r15))]
elif x==6:
IR16_cnt=IR16_cnt+1
f_cond_r16.append(a)
f_pow_r16.append(b)
f_mom_r16.append(c)
f_ct[x]=[dict(Counter(f_cond_r16)),dict(Counter(f_pow_r16)), dict(Counter(f_mom_r16))]
elif x==7:
IR17_cnt=IR17_cnt+1
f_cond_r17.append(a)
f_pow_r17.append(b)
f_mom_r17.append(c)
f_ct[x]=[dict(Counter(f_cond_r17)),dict(Counter(f_pow_r17)), dict(Counter(f_mom_r17))]
elif x==8:
IR18_cnt=IR18_cnt+1
f_cond_r18.append(a)
f_pow_r18.append(b)
f_mom_r18.append(c)
f_ct[x]=[dict(Counter(f_cond_r18)),dict(Counter(f_pow_r18)), dict(Counter(f_mom_r18))]
elif x==9:
IR19_cnt=IR19_cnt+1
f_cond_r19.append(a)
f_pow_r19.append(b)
f_mom_r19.append(c)
f_ct[x]=[dict(Counter(f_cond_r19)),dict(Counter(f_pow_r19)), dict(Counter(f_mom_r19))]
elif x==10:
IR20_cnt=IR20_cnt+1
f_cond_r20.append(a)
f_pow_r20.append(b)
f_mom_r20.append(c)
f_ct[x]=[dict(Counter(f_cond_r20)),dict(Counter(f_pow_r20)), dict(Counter(f_mom_r20))]
elif x==11:
f_cond_r21.append(a)
f_pow_r21.append(b)
f_mom_r21.append(c)
IR21_cnt=IR21_cnt+1
f_ct[x]=[dict(Counter(f_cond_r21)),dict(Counter(f_pow_r21)), dict(Counter(f_mom_r21))]
elif x==12:
f_cond_r22.append(a)
f_pow_r22.append(b)
f_mom_r22.append(c)
IR22_cnt=IR22_cnt+1
f_ct[x]=[dict(Counter(f_cond_r22)),dict(Counter(f_pow_r22)), dict(Counter(f_mom_r22))]
elif x==13:
f_cond_r23.append(a)
f_pow_r23.append(b)
f_mom_r23.append(c)
IR23_cnt=IR23_cnt+1
f_ct[x]=[dict(Counter(f_cond_r23)),dict(Counter(f_pow_r23)), dict(Counter(f_mom_r23))]
elif x==14:
f_cond_r24.append(a)
f_pow_r24.append(b)
f_mom_r24.append(c)
IR24_cnt=IR24_cnt+1
f_ct[x]=[dict(Counter(f_cond_r24)),dict(Counter(f_pow_r24)), dict(Counter(f_mom_r24))]
elif x==15:
f_cond_r25.append(a)
f_pow_r25.append(b)
f_mom_r25.append(c)
IR25_cnt=IR25_cnt+1
f_ct[x]=[dict(Counter(f_cond_r25)),dict(Counter(f_pow_r25)), dict(Counter(f_mom_r25))]
for x in range(0, 16):
for y in range(0,3):
if y==0:
print("ring=", x+10, "cond", f_ct[x,y])
elif y==1:
print("ring=", x+10, "pow", f_ct[x,y])
elif y==2:
print("ring=", x+10, "mom", f_ct[x,y])
print("")
df = pd.DataFrame.from_dict({'Ring 10 Cond': pd.Series(f_ct[0,0]), 'Ring 10 Pow': pd.Series(f_ct[0,1]), 'Ring 10 mom': pd.Series(f_ct[0,2])})
ax = df.plot.bar()
labs = [round(float(t.get_text()), 1) for t in ax.axes.get_xmajorticklabels()]
ax.set_xticklabels(labs)
plt.xlabel("Ring")
plt.ylabel("Matches")
plt.show()
f_ct字典包含15行3列,每个索引包含一个0.1-0.9之间的“匹配”计数器。我试图将每一行在x轴上绘制为一个“环”,并在y轴上为每个值0-0.9绘制“匹配”的数量。我想做3个图,字典中每个y索引(cond、pow和mom)对应一个图。Fu ct的打印也是如此
环=10秒{0.1:1,0.2:1,0.3000000000004:1,0.4:1,0.5:1,0.6:1,0.700000000000001:1,0.8:1,0.9:1}
环=10波{0.9:9}
环=10个月{0.9:9}
环=11秒{0.1:4,0.2:4,0.3000000000004:4,0.4:4,0.5:4,0.6:4,0.700000000000001:4,0.8:4,0.9:4}
环=11波{0.8:18,0.9:18}
环=11个月{0.8:18,0.9:18}
等等
该图在x轴上显示为0.1-0.9,而不是“环”号。我希望每个值0.1-0.9在每个“环”上都有自己的列,并由计数器指示“匹配”的数量。因此,第二个图“环10”的每个值0.1-0.9都有1个“匹配”,而“环11”的每个值0.1-0.9都有4个“匹配”。而pow图将在“第10环”上有9个“匹配”表示0.9,在0.1-0.8之间没有匹配,“第11环”将有18个“匹配”表示0.8,18个表示0.9,在0.1-0.7之间没有匹配。等等
谢谢大家!
目前没有回答
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